Strength of Materials Virtual Lab

1

Strength of Materials Virtual Lab 2021

1. TENSILE TESTING OF MATERIALS

Purposes and Objectives of Laboratory Work

Experimental determination of mechanical characteristics: yield strength σf, tensile

strength σt, true tensile strength Sk, elongation δ and relative narrowing ψ of the material sample

after rupture.

Laboratory Equipment Description Testing machines are used to determine the mechanical characteristics of materials.

Explosive and universal testing machines of all systems are used. The appearance of the testing

machine with an electromechanical exciter is shown in Figure 1.1.

Figure 1.1 – Laboratory Equipment General View:

1 – Test Sample; 2 – Dynamometer; 3 – Grips; 4 – Traverse; 5 – Laptop

Brief Theoretical Information

In a tensile test, a specimen of a certain shape and size from the test material is firmly

fixed with its ends (heads) in the grips of the testing machine and undergoes continuous smooth

deformation until fracture. In this case, the relationship between the tensile load and the

elongation of the design part of the sample is recorded in the form of a sample tension diagram.

For tensile tests, seven types of standard are used. One of the types is shown in Figure 1.2.

Figure 1.2 – The Geometric Model of the Sample:

d0 – Diameter of the Design Part of the Sample; l0 – Length of the Design Part of the Sample;

l – Length of the Working Part of the Sample.

l

l0

d0

2


The ratio of l0 to d0 must be strictly defined. The standards provide 100

0 d

l or 5

0

0 d

l.

During testing, the following basic conditions must be met: high-quality centering of the

sample in the grips of the testing machine, smooth deformation, the speed of movement of the

active grip during the test to the yield strength of not more than 0.1, beyond the yield strength of

not more than 0.4 of the length of the design part of the sample per minute, the ability to suspend

loading with an accuracy of one smallest division of the scale of the load meter, smooth

unloading.

Tensile diagrams of samples of low carbon steel (С≤0.3%), constructional steel

(С≥0.35%) and gray cast iron are shown in Figures 1.3 a, 1.3 b and 1.3 c.

Figure 1.3 – Stretch Diagrams of Samples:

Mild Steel (a), Constructional Steel (b), Cast Iron (c)

Let us consider in more detail the tensile diagram of a sample of low carbon steel (Figure

1.4). In the initial section of the diagram, between the force F and the elongation ∆l, a direct

proportional dependence is observed – the sample obeys Hooke's law.

Figure 1.4 – Mild Steel Sample Tensile Diagram

At some point in the diagram, Hooke's law is violated: the relationship between force and

elongation becomes nonlinear. In the diagram, there is a horizontal area called the «yield area».

At this stage of the test, the sample is elongated (deformed) at almost constant force. This

Δlp Δle

Δl

Δl

F

E D

C B

А* А

0

F

Re-loading

Unloading

Break

Unloading

Ff

Fe

Fp

r

a) b) c)

F

Δl

F

Δl Δl

F

3


phenomenon is called «yield», and the sample is deformed evenly along the entire length of its

working part. Then the «yield area» ends and the «hardening area» begins. At the end point of

this section, the maximum force that the sample can withstand is reached. Then begins the «site

of destruction» or «site of local fluidity». A local thinning (neck) appears on the sample. The

neck diameter decreases as the sample deforms, and the sample breaks along the smallest neck

section.

If during a tensile test the load is suspended, for example, at point D of the diagram

(Figure 1.4) and the sample is unloaded, then it turns out that the unloading diagram and the

previous loading diagram do not coincide. The discharge line is a straight line parallel to the

initial linear portion of the sample stretch diagram. This nature of the deformation of the sample

during its unloading is called the law of unloading.

Upon repeated loading, the diagram to point D coincides with the unloading line, and

then it will coincide with the tensile diagram of the specimen under a single loading. This type of

deformation is called the law of reloading and consists in a direct proportional dependence of the

force and elongation, which is maintained up to the value of the force achieved during initial

loading.

When unloading a sample within the ОА* section, the laws of loading, unloading and

reloading coincide.

Elongation of the sample during its deformation beyond the point A* diagrams consists of

elastic and plastic elongations (Figure 1.4), i.e.

pe lll , (1.1)

where ∆l, ∆le, ∆lp – full, elastic and plastic elongation of the calculated part of the sample.

When unloading a sample that has received elastic and plastic elongations, the elastic

elongation decreases, in accordance with the law of unloading, and the plastic remains

unchanged.

The tensile diagram of the sample allows you to evaluate the behavior of the material of

the sample in the elastic and elasto-plastic stages of deformation, as well as determine the

mechanical characteristics of the material.

To obtain numerically comparable mechanical characteristics of materials, the tensile

diagrams of the samples are rearranged into the tensile diagrams of the materials, i.e. depending

on the stress σ and strain ε, which are determined by the formulas

0A

F , (1.2)

0l

l , (1.3)

where F – the force acting on the sample; A0, l0 – the initial cross-sectional area and the initial

length of the calculated part of the sample.

The tensile diagram of the material obtained under these conditions (without taking into

account the change in the size of the calculated part of the sample) is called the conditional

tensile diagram of the material, in contrast to the actual tensile diagram, which is obtained taking

into account changes in the size of the sample. The tensile diagram of a material depends on its

structure, test conditions (temperature, strain rate).

For mild steel, the tensile diagram is shown in Figure 1.5.

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Figure 1.5 – Material Tensile Diagram (Low Carbon Steel)

Within the plot of the OA diagram, Hooke's law is observed, i.e.

E (1.4)

The proportionality coefficient E is called the elastic modulus of the first kind, or

Young's modulus. It characterizes the resistance of the material to elastic deformation. This value

is the constant elasticity of the material.

Hooke's law is violated at point A of the diagram. The ordinate of this point has a special

name – the limit of proportionality.

Figure 1.6 – Determination of the Conditional Limit of Proportionality

L M K

A

a

a

σ

0 ε

σpr

β

α

εp ε

F

E D

C B

А* A

0

σ

σf

σe

σp

r

εe

ε εdes

σt

True

Diagram

Conditional

Diagram

σd

es

5


The proportionality limit σpr is the greatest stress to which Hooke's law is valid. It is

essentially impossible to use this definition of the limit of proportionality for the practical

calculation of its value. Therefore, the concept of a conditional (technical) limit of

proportionality is introduced. It is estimated as the stress at which the deviation from the direct

proportional relationship between stress and strain reaches a certain value.

The conditional (technical) limit of proportionality is the stress σpr at which the tangent of

the slope of the tangent to the curve σ = f(ε) is 1.5 times the tangent of the slope of the linear

section of this diagram (the angle is counted from the axis σ) (Figure 1.6).

With a certain excess of the proportionality limit, all deformations continue to remain

elastic, i.e. completely disappearing if the stress is reduced to zero. The highest stress to which

all strains in the material are elastic is called the elastic limit σe. In practice, use the conditional

elastic limit.

The conditional (technical) elastic limit σ0.01 is the stress at which the residual (plastic)

deformation is 0.01%. To find point A* (Figure 1.7) corresponding to the elastic limit, it is

necessary to use the law of unloading and reloading.

Figure 1.7 – Determination of Conditional Elastic Limit

The yield point σf , called the yield point, corresponds to the yield point of the tensile

diagram of low carbon steel. The yield stress (physical) σf is the stress at which residual (plastic)

strains are intensively accumulated in the material, and this process occurs at an almost constant

stress.

In the absence of a yield point (tensile diagrams of most materials), the conditional

(technical) yield strength is determined. The conditional (technical) yield stress σ0.2 is the stress

at which the residual (plastic) strain is 0.2%.

The conditional yield strength is determined similarly to the elastic limit (Figure 1.8).

σ

ε

A

A*

εp=0.0001 0

σ0.01

6


Figure 1.8 – Determination of Conditional Yield Strength

The CDE section of the material tensile diagram (Figure 1.5) has a maximum at point E.

The ordinate of this point is called the conditional tensile strength (temporary resistance) and is

defined as:

0

max

tA

F (1.5)

For materials in a plastic state under these conditions, the tensile strength σt is not equal

to the actual stress in the sample material, since by the time Fmax is reached, the cross-sectional

area of the sample is significantly reduced. Before the formation of the neck (point D of the

diagram), the deformation of the calculated part of the sample is uniform and consists of elastic

(reversible) and plastic (residual) (Figure 1.5).

The stress state before the formation of the neck is uniaxial: in areas coinciding with the

cross section, there are normal tensile stresses; in all areas perpendicular to the cross section, the

stresses are zero. It is essential that the stresses at all points of the same section are the same, and

the internal forces at all sections on the working section are equal. In the final section of the

deformation (after the occurrence of the neck), localization of deformations in the neck occurs;

in the rest of the sample, it practically does not increase. Deformation in the neck is

heterogeneous, has a high gradient along the axis of the sample. The stress state also becomes

inhomogeneous, in addition, it changes qualitatively – it becomes triaxial. Inside the neck, the

stress state is triaxial tension.

The sample is torn along the smallest cross section of the neck (Figure 1.9) at a stress

significantly exceeding the tensile strength. This stress is called true tensile strength:

min

desk

A

FS , (1.6)

where Fdes – the force at which the sample breaks; Amin – the cross-sectional area of the neck at

the rupture:

4

2

minmin

dA

(1.7)

The plastic properties of materials are evaluated by two characteristics: elongation after

rupture and relative narrowing after rupture.

Elongation after rupture δ:

A

A*

B

σ0.2

εp = 0.002 ε 0

σ

7


%1000

0after

l

ll , (1.8)

where lafter – the length of the design part of the sample after rupture, l – the length of the design

part of the sample before the test.

Relative narrowing after rupture ψ:

%1000

min0

A

AA , (1.9)

where A0 – the cross-sectional area of the estimated sample length before testing.

Figure 1.9 – Destroyed Sample

The methodology for determining the mechanical characteristics of materials having

tensile diagrams other than that of mild steel remains unchanged.

Machine Diagram Processing

The tensile diagram of the sample displayed on the PC screen is called the machine

tensile diagram of the sample and contains a significant error caused by the method of measuring

the elongation of the working part of the sample. The elongation of the working part of the

sample is measured by the movement of the movable yoke of the test machine. To determine the

mechanical characteristics of a material, the converted machine diagram is usually represented in

two parts. First, the diagram is displayed completely from the beginning of deformation of the

sample to its destruction, after which the diagram is displayed from the beginning of deformation

to the end of the yield point.

The sequence of processing the machine diagram of the tensile sample:

1. Continue the linear (initial) section of the diagram to the upper and lower boundaries

of the grid (points L and O).

2. Calculate the scale of the diagram along the F and ∆l (MF, N/mm; M∆l, mm/mm).

3. Determine the elastic elongation of the working part of the sample for the force F0 by

the formula

0

00

AE

lFl

, (1.10)

where F0, N – the force value corresponding to the upper boundary of the coordinate grid; l, mm

– the distance between the heads of the sample before the test; E – the modulus of elasticity

(E=2ˑ105 for steel); A0 – the initial cross-sectional area of the sample:

4

2

00

dA

(1.11)

4. Put the segment Δl0 on the scale M∆l from point L to the left (point N).

5. Connect the points O and N of the line. The total elongations of the working part of the

sample are counted from this straight line.

dmin

lafter

8


6. Perform similar operations for the full chart..

The Order of Execution of the Laboratory Work

1. Measure the diameter d0 of the working part of the test sample. Длину Take the length

of the calculated part of the sample l0 equal to the length of the working part of the sample l, i.e.

equal to the distance between the heads of the sample. A change in this particular size is

recorded on the diagram by the testing machine as ∆l .

2. Run a test and get a machine diagram.

3. Measure the destroyed sample (neck diameter dmin and the final length of the working

part lafter). To assess the final length of the working part, connect the parts of the destroyed

sample and measure the distance between its heads.

4. Play the resulting machine deformation diagram on two sheets of A4 format.

Determine the scales along the axes of the machine diagram.

5. Process the machine diagram in accordance with the diagram presented earlier.

Mark six points on the diagram (A, B, C, D, E, F):

- point A corresponds to the conditional limit of proportionality;

- point B corresponds to the yield strength (physical or conditional);

- point C corresponds to the end of the yield site;

- point D – intermediate between points C and E;

- point E corresponds to the maximum force Fmax;

- point F corresponds to the break of the sample.

6. Record the processing results in the laboratory journal. To present the results of the

experiment, it is recommended to use tables 1.1–1.3. As the yield strength and tensile strength,

take stresses corresponding to the points. B and E diagrams (Figure 1.5).

7. Based on the results of processing and calculations, build a tensile diagram of

materials.

Table 1.1 – The Geometric Dimensions of the Samples Before and After the Tensile Test

Material Mild Steel Constructional

Steel Gray Cast Iron

Diameter of the sample before

testing d0, mm

Length of the calculated part of the

sample before the test l0, mm

Cross-sectional area of the sample

before testing, A0, mm2

Diameter of the sample after testing

dafter, mm

Length of the calculated part of the

sample after testing lafter, mm


after the test, Aafter, mm2

9


Table 1.2 – Tensile Test Results

Point Mild Steel Constructional Steel Gray Cast Iron

F, kN Δl, mm σ, MPa ε F, kN Δl, mm σ, MPa ε F, kN Δl, mm σ, MPa ε

A

B

C

D

E

F

Table 1.3 – Mechanical Characteristics of the Tested Materials

Material σpr, MPa σf, MPa σe, MPa σt, MPa Sk, MPa δ, % ψ, %

Mild Steel

Constructional Steel

Gray Cast Iron

Control Questions

1. What kind of tensile diagrams do samples of low carbon steel, structural steel and cast

iron have?

2. How to determine the amount of elastic/residual elongation corresponding to a given

load from the tensile diagram of the sample?

3. What is called the physical and conditional limit of proportionality?

4. What mechanical characteristics determine the ability of a material to plastically

deform?

5. How do the lines of intermediate unloading and reloading appear on the tensile

diagram of the sample?

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2. COMPRESSION TESTING OF MATERIALS


Experimental determination of mechanical characteristics: yield strength σf of low-carbon

steel and tensile strength σt of gray cast iron under compression.

Laboratory Equipment Description

For compression tests, universal testing machines that meet the requirements of the

standard are used (Figure 2.1). The conditions that must be met during the compression test are

the same as in the tensile test, but higher demands are placed on the centering of the sample and

the absence of mutual misalignment of the pressure plates transmitting force to the sample. The

compression test is widely used to determine the mechanical characteristics of low-plastic

materials, for example, cast irons, tool steels, ceramics, etc.


1 – Test Sample; 2 – Dynamometer; 3 – Bottom Plate; 4 – Top Plate;

5 – Movable Traverse; 6 – Laptop


In a compression test, a sample of standard shape and size from the test material is placed

in a fixture mounted on a testing machine and undergoes continuous, smooth deformation to a

predetermined strain value or to failure. In this case, the relationship between the compressive

force F and the shortening of the height h of the sample is recorded in the form of a compression

diagram of the sample. The compression diagram of the sample allows you to evaluate the

behavior of the material of the sample in the elastic and elastoplastic stages of deformation and

determine the characteristics of the mechanical properties of the material.

Compression test of material samples is carried out according to the standard. Four types

of cylindrical specimens are used: three types of specimens with smooth ends and one with

recesses at the ends. The type of samples is selected depending on the characteristics to be

determined. For compression tests, as a rule, short samples with a height to diameter ratio of 1–3

11


are used (Figure 2.2). The use of high samples is unacceptable, because such samples will not

only shrink, but also bend.

Figure 2.2 – Geometric Model of the Sample

Samples must be carefully prepared, in particular, tight tolerances on the perpendicularity

of the axis of the sample to its ends are observed. The ends of the sample must be carefully

ground.

The compression test has characteristic features that significantly distinguish it from a

tensile test:

- samples of plastic materials are not destroyed, receiving a significant deformation

superior to the deformation at break under tensile conditions;

- the compression test results of the samples significantly depend on the ratio of the

height of the sample to its diameter;

- the tensile strength and ductility characteristics are markedly affected by the friction

forces at the supporting ends of the sample.

In the process of loading the sample with compressive forces, its height decreases and the

diameter increases, and its diameter increases unevenly along the height of the sample. This

leads to a significant change in shape - the sample becomes barrel-shaped. Barrel compression

occurs due to friction between the contacting surfaces of the compressible sample and device.

Friction prevents lateral deformation at the ends of the sample (Figure 2.3).

Figure 2.3 – Changing the Shape of the Sample During Compression

F

h

d0

12


The stress state in the sample, with developed barrel-shaped, is not uniaxial and

inhomogeneous. It is not possible to take this heterogeneity into account when processing the

results of compression tests; therefore, it is assumed that the stress state is uniform and uniaxial

over the entire volume of the sample. Thus

0A

F , (2.1)

where σ – the normal stress in the cross section, F – the force acting on the sample, A0 – the

initial cross sectional area.

In order for the actual stress state in the sample to correspond to the expected one, it is

necessary to reduce or eliminate the friction forces at the ends of the sample. This is achieved by

introducing lubricant at the ends of the sample (Figure 2.4) or by creating conical end surfaces

with an angle α equal to the angle of friction between the materials of the sample and the device

(Figure 2.5).

Figure 2.4 – Sample with End Cavities

You can combine the above methods of struggle with the forces of friction. As a

lubricant, paraffin, waxed paper, petroleum jelly, teflon, etc. are used.

Figure 2.5 – Samples with Conical Ends:

with a Central Hole (a); without Hole (b)

It is not possible to completely eliminate the frictional forces between the contacting

surfaces during the compression tests. This is the fundamental disadvantage of these tests.

α α

a) b)

Cavity for Lubricant

13


The smaller the ratio of the height of the sample to its diameter, the greater the influence

of friction on the test results. From these positions, tests should be carried out using as long

samples as possible. However, when compressing long samples, it is difficult to avoid bending.

The optimal ratio for cylindrical samples is h/d0=1...3.

A material compression diagram is obtained in the same manner as a tensile diagram. The

method for determining the mechanical characteristics of a material, such as the proportionality

limit, the elastic limit, and the yield strength, is fully consistent with the method for determining

these characteristics in a tensile test.

It is not possible to establish the tensile strength σt of low-carbon steel under

compression, since a sample of this material is flattened, remaining solid, i.e. not destroyed.

The compression chart of the material is obtained from the compression chart of the

sample, while taking

0A

F , (2.2)

0h

h (2.3)

The compression diagram of low-carbon steel (Figure 2.6 a) in the area of elastic

deformations and general yield almost coincides with the tensile diagram of this material. There

is no point on the compression diagram corresponding to the tensile strength; therefore, the

tensile strength cannot be established. Compression tests of a material in a plastic state are

stopped at a strain of approximately 50%.

Figure 2.6 – Sample Compression Diagrams:

Mild Steel (a); Constructional Steel (b); Cast Iron (c)

The conditional limits of proportionality σpr, elasticity σ0.01, yield stress σ0.2 are

determined by the same method as in tension.

Under normal conditions, gray cast iron is in a slightly plastic state; the compression

diagram of gray cast iron is shown in Figure 2.6 c. Strictly speaking, there is no linear section on

the tension and compression diagrams of gray cast iron where Hooke's law is observed, however,

the deviation from Hooke's law is small.

F F F

Δh Δh Δh a) b) c)

14


Unlike mild steel, cast iron is destroyed by compression. Failure occurs along sites tilted

at an angle of 450 to the axis of the specimen, where the greatest tangential stresses occur.


1. Place steel sample №1 between the pressure plates of the device, carefully centering it.

2. Gently load the sample with compressive force.

3. On the machine compression diagram, note the force corresponding to the beginning of

the fluidity of the material.

4. Continue loading the sample to a noticeable barrel shape (cylindrical distortion).

5. Unload the sample and remove it from the device.

6. Install steel sample №2 by placing paraffin gaskets between the ends of the sample and

the pressure plates of the device.

7. Repeat steps 2–5. The maximum load should be the same as for sample №1.

8. Install a sample of cast iron in the fixture.

9. Gently load the sample with compressive force until fracture. Note the destructive

force in the machine compression diagram.

10. Unload the sample and remove it from the device.

11. Find the yield strength of steel, the tensile strength of cast iron.

12. Based on the results of processing and calculations, construct a diagram of the

compression of materials.

To present the results of the experiment, it is recommended to use tables 2.1–2.3.

Table 2.1 – The Geometric Dimensions of the Samples Before and After the Compression Test

Test Smaple

Steel

(without greasing

the ends)

Steel

(with paraffin

pads)

Cast Iron

Diameter of the sample before

testing d0, mm

Sample height before test

h0, mm


before testing, A0, mm2

Diameter of the sample after testing

dafter, mm

Sample height after test hafter, мм


after testing, Aafter, mm2

15


Table 2.2 – Compression Test Results

Point

Steel

(without greasing the ends)

Steel

(with paraffin pads) Cast Iron

F, kN Δh,mm σ, MPa ε F, kN Δh,mm σ, MPa ε F, kN Δh,mm σ, MPa ε

A

B

C

D

E

F

Table 2.3 – Mechanical Characteristics of the Tested Materials

Material σpr, MPa σf, MPa σe, MPa σt, MPa

Steel

Cast Iron

Control Questions

1. What are the features of deformation during compression of samples from plastic and

brittle materials?

2. What mechanical strength characteristics are obtained during testing of brittle

material?

3. For some materials, the mechanical characteristics of tensile and compression strength

are almost the same, for which they are different?

4. What are the main compression tests for which materials?

5. Describe how to perform compression testing of material samples.

16


3. TORSION TESTING OF MATERIALS


Experimental determination of the mechanical characteristics under pure shear: shear

modulus G, yield strength τf, tensile strength τt.


In a torsion test, a sample of the material under study is firmly fixed by the heads in the

grips of the testing machine and undergoes continuous smooth deformation until fracture. For

torsion tests, any test machine that meets the requirements of the standards can be used. The test

machine diagram is shown in Figure 3.1.

Figure 3.2 – Scheme and General View of the Laboratory Equipment:

1 – Test Sample; 2 – Loading Mechanism; 3 – Grips;

4 – Protective Screen; 5 – Moment Meter; 6 – Laptop

Electric Motor

Angle Meter

Movable

Carriage

Construction

Gearbox

17



When testing samples for torsion, the following basic conditions must be met: high-

quality centering of the sample in the grips, smooth loading and unloading, and the absence of

longitudinal force.

For torsion tests, standard samples are used. Both solid and tubular samples can be used.

A solid cross-sectional sample is shown in Figure 3.2.

Figure 3.2 – Test Sample

The design of the sample heads should ensure the transmission of torque from the active

capture to the working part of the sample and from the working part to the torque meter.

Dimentions of the sample: diameter of the design part d0, length of the design part l0, length of

the working part l, in this case, 00 dll and 100

0 d

l or 5

0

0 d

l must be observed.

The torsion test is carried out both for plastic and for low-plastic and brittle materials. In

the torsion test, a neck does not form on the samples, as a result of which the torque increases

until the sample breaks. Plastic deformation is uniform along the length of the sample, the

cylindrical shape of the solid sample is maintained until fracture.

Consider the torsion diagram of a sample of plastic material (Figure 3.3 a).

Figure 3.3 – Torsion Diagrams of Samples:

Mild Steel (a); Gray Cast Iron (b)

In the initial section of the diagram, between the torque Mt and the twist angle φ, a direct

proportional dependence is observed – the sample is deformed linearly-elastic. Then begins the

«hardening section», characterized by a low rate of increase in torque and a large increment of

the angle of rotation of the twist. The end point of the diagram corresponds to the destruction of

the sample. Samples of different materials during torsion testing break down in different ways:

the fracture surface perpendicular to the axis of the specimen indicates failure from shear, and

the helical surface indicates fracture from separation.

Мt

0 φ

Мt

0 φ a) b)

l

l0

d0

18


The twist angle of the sample during its deformation outside the elastic section of the

diagram is the sum of the angle φe corresponding to the elastic shear strain, obeying the Hooke

law, and the twist angle φp corresponding to the plastic shear strain, i.e. φ=φe+φp.

The torsion diagram of the sample allows you to evaluate the behavior of the material of

the sample in the elastic and elastoplastic stages of deformation, to determine the mechanical

characteristics of the material under shear. To obtain numerically comparable mechanical

characteristics of materials, the torsion diagrams of samples are rearranged into shear diagrams

of materials. The material shear diagram refers to the relationship between the shear stresses τ,

occurring in the material during shear and the corresponding shear angles γ.

The shear stresses in the sample of circular cross section within the elastic range are

proportional to the distance of the point from the axis of the sample:

rI

M

P

t , (3.1)

where Mt – the torque, IP – the circular moment of inertia of the circular cross section, r – the

distance of the point from the axis of the sample.

The greatest tangential stresses during torsion of a circular cross-sectional specimen

occur at the cross-sectional points at the outer cylindrical surface and, within the limits of

applicability of Hooke's law, are calculated by the formula:

PW

M tmax , (3.2)

where WP – the moment of torsion resistance of circular cross section.

The material shear diagram is constructed from the torsion diagram of the sample.

In the initial part, the shear diagram is linear, i.e. the shear stress τ is proportional to the

shear angle γ. The law of proportionality, called the Hooke law under shear, can be written in the

form:

G , (3.3)

where the proportionality coefficient G is called the shear modulus or elastic modulus of the

second kind. It characterizes the resistance of the material to elastic deformations and is an

elastic constant. Note that the three elastic constants (Young's modulus E, shear modulus G, and

Poisson's ratio ν) are not independent, but are related by the relation:

v

EG

12 (3.4)

Hooke's law during shear is satisfied up to a stress called the proportionality limit τpr.

To determine the shear modulus G, we use the dependence of the twist angle φ on the

torque Мt:

PIG

aM

t , (3.5)

where a – the distance between the sections of the sample, the mutual rotation angle of which is

measured.

From here:

PI

aMG

t (3.6)

By measuring the twisting angles φ corresponding to the values of the torque Mt, we can

calculate the shear modulus G.

Attention should be paid to the different nature of fracture under torque loading of

samples made of ductile (low-carbon steel) and brittle (cast iron) material. A plastic specimen is

destroyed along the cross section due to shear stresses («shear»), and a brittle specimen is

destroyed along a helical surface inclined to the axis of the specimen at an angle of 450, where

the highest tensile stresses act («separation»).

19



1. Measure and record the diameter of the working part and the working length of steel

and cast iron samples.

2. Install a steel specimen in the machine grips.

3. Follow the procedure of loading the sample to its destruction.

4. Fix the value of the breaking moment.

5. If necessary, reproduce the resulting graph on a sheet of A4 paper.

6. Install a cast-iron sample in the grips of the machine.

7. Repeat steps 3-5 for a cast iron sample.

8. Draw sketches of the destroyed samples and build torsion diagrams of steel and cast

iron.

To present the results of the experiment, it is recommended to use table 3.1.

Table 3.1 – Torsion Test Results

Diagram Point

Steel Sample Cast Iron Sample

Torque Мt, N·m Twist Angle φ Torque Мt, N·m Twist Angle φ

Control Questions 1. What is the purpose of torsion testing of material samples?

2. What samples are used for torsion testing?

3. Describe the principle of operation of the testing machine for testing samples of

materials for torsion.

4. What is the fundamental difference between the torsion diagrams of steel and cast iron

samples?

5. What is the torsional yield strength?

20


4. ELASTIC CONSTANTS OF ISOTROPIC MATERIALS


Experimental determination of the elastic modulus E and Poisson's ratio v of an isotropic

material.


A rectangular cross-section rod (cross-sectional dimensions: 50×3 mm), mounted in the

grips of the testing machine (Figure 4.1) and loaded with a longitudinal force F, is used as the

test object.


1 – Test Rod; 2 – Dynamometer, 3 – Upper Grip;

4 – Lower Grip; 5 – Strain Gauges; 6 – Laptop; 7 – Calibration Beam;

8 – Watch Type Indicator (Kf=0.01); 9 – Weights; 10 – Meter of Deformations


Prior to testing, two strain gages are glued to the sample (Figure 4.2): one in the

longitudinal direction (for measuring longitudinal strain εz) and the other in the transverse

direction (for measuring transversal strain εx). Both foil type strain gages. The strain gauge

circuit is shown in Figure 4.3.

21


Figure 4.2 – The Location of the Strain Gauges on the Rod:

1 – Rod, 2 – Transverse Strain Gauge, 3 – Longitudinal Strain Gauge

Figure 4.3 – Strain Gauge Scheme:

1 – Substrate, 2 – Strain-Sensing Element, 3 – Output Conductors

A schematic diagram of connecting strain gauges to a strain gauge is shown in Figure 4.4.

Temperature compensation is carried out by including in one of the arms of the bridge circuit a

compensation strain gauge (R2), which is glued to a bar of the same material as the test sample,

but is not loaded during the test.

1 2 3

y

x

z F

F

a

b

1

2

3

22


Figure 4.4 – Scheme of a Device for Measuring Strain Using Strain Gauges:

R1 – Resistance of the Working Strain Gage; R2 – Resistance of the Compensation Strain Gage;

R3, R4 – Resistances Built into the Device;

S – Slidewire Scale; G – Galvanometer

The laboratory unit is equipped with a device for calibrating a strain gauge (Figure 4.5).

Figure 4.5 – Device for Calibrating the Deformations Meter Scale Division:

1 – Beam, 2 – Strain Gage, 3 – Deflection Meter

Hooke's law determines the proportionality between stresses and elastic strains:

zz E , (4.1)

r

h

F

1 2 3

2

z

b(z)

l

R1 R2

R3 R4

G

Rr

S

23


where E – the elastic modulus of the first kind.

Hooke's law is valid up to stress σpr, called the material proportionality limit.

Under axial loading of the rod, not only longitudinal (σz), but also transverse

deformations (σx, σy) take place.

When the sample is loaded to the limit of proportionality of the material, the ratio

between the transverse and longitudinal deformations is constant. The absolute value of the ratio

of transverse to longitudinal strain is called the Poisson's ratio or transverse strain coefficient:

z

xv

long

trans (4.2)

The elastic modulus and Poisson's ratio completely determine the elastic properties of an

isotropic material.

The elastic constants of the material can be determined experimentally by dynamic and

static methods. Currently, it is believed that the dynamic method allows to obtain elastic

constants with higher accuracy than the static method. The dynamic method is based on the

dependence of the oscillation frequency of the test sample (resonance methods) or the speed of

sound in it (pulsed methods) on the elastic constants. The static method involves smooth loading

of a material sample with simultaneous measurement of longitudinal and transverse strains.

To measure deformations, instruments called tensometers are used. Various types of

strain gauges have been developed and applied: mechanical, optical, pneumatic, electrical and

others. The most common are electric strain gauges. Their action is based on a change in the

parameters of the electric circuit of the strain gauge (resistance, capacitance or inductance) in

accordance with the measured strain. In the present experimental work, electric resistance

tensometers are used. The resistance strain gauge consists of two parts: a strain gauge (figure

4.3) and a recording device (electronic deformation meter).

Strain gages adhere firmly to the outer surface of the sample. When the sample is

deformed, strain gages are also deformed, as a result of which their electrical resistance changes.

Widespread wire and foil strain gauges. Strain-sensitive elements of wire strain gages are

made in the form of a loop-shaped lattice from a thin (diameter 0.01-0.03 mm) wire. Strain-

sensitive elements of foil strain gages are made of thin foil (thickness 0.005–0.01 mm). Foil

strain gauges in comparison with wire have better metrological characteristics. The materials

used for the strain-sensitive element are: constantan (Cu 58.5%, Ni 40%, Mn 1.5%), elinvar (Ni

36%, Cr 8%, Fe 56%), nichrome (Ni 80%, Cr 20% ) and other alloys.

Structurally, strain gages (Figure 4.3) consist of a varnish, paper or metal substrate on

which a strain-sensitive element is fixed with glue. The strain gauge is connected to the strain

gauge using output conductors soldered to it.

An important characteristic of a strain gage is its base. Currently, wire strain gages with a

base of 3 to 50 mm are used, foil strain gages have a base of 0.3 to 100 mm. The use of strain

gauges of one or another base is associated with the nature of the deformed state of the

investigated object. The higher the strain gradient, the smaller should be the base of the strain

gauges used.

The strain gauge (Figure 4.4) is most often assembled using a bridge circuit. When using

strain gages, the issue of temperature compensation is important, since a change in temperature

causes a change in the electrical resistance of the strain gage. Temperature compensation is

carried out by introducing a compensation strain gage into the corresponding arm of the bridge

circuit. The compensation strain gage is glued to a load-free bar of material of the same grade as

the material of the test object, and is placed in the same temperature conditions as the working

strain gage. The compensation strain gage is selected from the same batch as the worker.

Calibration of Electronic Deformation Meter

To determine the strain using a strain gauge, it is necessary to know the scale division

value of the deformation meter. The process of determining the price of dividing the scale of the

device is called its calibration. Calibration is carried out using special devices – calibration

24


f a

a / 2

1

2

3

beams (Figure 4.5). A beam of equal bending resistance is often used. This is a cantilever beam

having a wedge-shaped shape in plan with a constant thickness.

The curvature of the elastic line of the beam:

x

x

IE

M

1, (4.3)

where ρ – the radius of curvature of the arc of the elastic line of the beam, Mx – the bending

moment, E – Young's modulus, Ix – the moment of inertia of the section relative to the x axis.

A beam of equal bending resistance:

zPM x , (4.4)

12

22

12

33 htgz

bhI x

(α – wedge angle), (4.5)

const

htgE

P

3

2

121

(4.6)

i.e., the elastic line is a circle, so the deformations on the upper and lower surfaces of the beam

are the same in length.

Strain gages are glued onto the upper and lower surfaces of the wedge-shaped part of the

beam, which are necessarily selected from the same lot as the working strain gages, i.e. glued to

the test sample.

When calibrating, the strain εz is determined by two independent methods – an electric

tensometer and a mechanical device (deflection meter). Comparing the readings of the

electrotensometer with the strain value measured by a mechanical device, one can find the

division price of the scale of the strain gauge.

The strain εz can be determined using a mechanical deflection meter, a diagram of which

is shown in Figure 4.6. Directly the deflection meter allows you to measure the deflection arrow

f of the calibration beam at the base a. It is necessary to connect the arrow of the deflection f with

the strain εz of the upper surface of the beam (Figure 4.7).

Figureк 4.6 – Deflection Meter Scheme:

1 – Beam; 2 – Deflection Base; 3 – Watch Type Indicator

25


Figure 4.7 – Scheme for Measuring the Deflection of the Calibration Beam

In Figure 4.7, the ОВС triangle is rectangular: ОС2 = ВС2 + ОВ2

2

2

2

2f

a

, (4.7)

024

22

ffa

(4.8)

The value of f 2 is small in comparison with other terms, it can be neglected.

Then

f

a

8

2

(4.9)

Deformation at the points of the upper face

2

2 h

d

ddh

z

, (4.10)

where from

2

4

a

hfz

(4.11)

On the other hand, the strain εz, measured by the strain gauge is

A

B C

f

a/2 a/2

ρ

O

Beam Elastic Line

26


nKz , (4.12)

where Kε – the division price of the scale of the strain gauge, nε – the number of divisions of the

scale of the information meter.

For the division price we get

2

4

an

hfK

(4.13)


1. Calibration of electronic deformation meter:

1.1. Connect strain gauge №3 mounted on the calibration beam to the strain gauge;

1.2. Load the calibration beam successively with forces P equal to 0, 20, 40 N, taking

each time the readings of the deflection meter and strain gauge.

1.3. Calculate the low-order price of a digital strain gauge, which is used in the main

experiment.

2. Determination of elastic modulus and Poisson's ratio:

2.1. Connect strain gauge №1 (longitudinal) mounted on the sample to the strain gauge.

2.2. Load the sample sequentially with forces of 5, 10, 15, 20 kN, taking each time the

readings of the deflection meter.

2.3. Unload the sample.

2.4. Connect strain gauge №2 (transverse) mounted on the sample to the strain gauge.

2.5. Repeat steps 2.2 and 2.3.

2.6. Calculate Poisson's ratio v and elastic modulus E.


Table 4.1 – Calibration Results of an Electronic Deformation Meter

Force P, N Deflection Readings Deformation Meter Readings

nf Δnf nε Δnε

0

20

40

ΔP = 20 N Average fn Average n

Deflection of the calibration beam per loading stage ff nKf

Linear deformation of the beam per loading stage 24 ahf

The scale division of the deformations meter nK

27


Table 4.2 – The Results of Determining the Elastic Modulus and Poisson's Ratio

Force P, kN

Deformations Meter Readings

Longitudinal Strain Gage Transverse Strain Gauge

nl Δnl nt Δnt

5

10

15

20

ΔP = 5 kN Average ln Average tn

Increments of longitudinal strain per loading step lzl nK

Increments of transverse strain per loading step txt nK

Poisson's ratio ltzxv

Normal stress increments per loading stage 0APz

Modulus of elasticity of the first kind (Young's modulus) zzE

Control Questions

1. Formulate Hooke's law under central tension-compression and give the formula for this

law.

2. What is Young's modulus and how is it measured?

3. What is the Poisson's ratio and what dimension does it have?

4. Describe the order of the experiment to determine Young's modulus and Poisson's

ratio.

5. What is a strain gauge?

28


5. DIRECT BENDING OF THE ROD


Experimental verification of the law of distribution of normal stresses in the cross section

of a rod with a clean bend. Determination of the displacement of the cross section of the rod,

comparison of experimental data with the calculation results.


The main element of the laboratory setup is: an I-section rod made of aluminum alloy,

mounted on two supports (Figure 5.1). The loading of the rod is carried out through the rocker

arm, which is simultaneously an elastic element of the force meter.


1 – I-beam; 2 – Loading Mechanism Screw; 3 – Rocker; 4 – Watch Type Indicators (Kf=0.01);

5 – Strain Gauges; 6 – Meter of Deformations (Kε=13.4·10–7)

l / 3 l / 3

l / 2

l

l / 4

A

I

I

29


The middle part of the rod (between the supports of the rocker arm) is in conditions of

clean bending. In the middle section (I-I) of the rod seven strain gages are glued (Figure 5.2) of a

foil type. Strain gages are installed in the direction of the longitudinal axis z and allow measuring

the strain εz at the corresponding points.

Figure 5.2 – The Strain Gauge Arrangement in Section I-I

Above one of the cross-sections of the rod, a deflection meter is installed, which is a dial-

type indicator mounted on a stand. The installation is completed with an electronic deformation

meter.


The technical theory of pure rod bending is based on the following hypotheses:

– the hypothesis of flat sections, according to which the cross-sections of the rod, flat

before deformation, remain flat after deformation.

– hypothesis of non-compressing longitudinal fibers, i.e. the longitudinal layers in the rod

with a clean bending of the rod do not interact in the direction perpendicular to them, therefore,

in areas parallel to the axis of the rod, the normal stresses are zero, therefore, the stress state with

a clean bending of the rod can be considered uniaxial.

When the rod bends, two zones are formed in it: the tension zone and the compression

zone. The boundary between the zones of tension and compression is a longitudinal layer called

neutral. This layer is curved, but its length does not change.

We introduce a rectangular coordinate system (x, y, z), where x and y are the central axes

of the section (the y axis lies in the plane of the bending moment, and the x axis lies in the

neutral layer).

Based on the hypothesis of flat sections, it can be concluded that the length of the

longitudinal layers changes: elongations Δl are directly proportional to y. The relative strains εz

also change, since the length of all the fibers of the rod before the strain is the same.

Stresses are associated with deformations by Hooke's law for a uniaxial stress state:

zz E (5.1)

where σz – normal stress in the cross section of the rod, E – elastic modulus of the first kind.

Theoretical values of normal stresses are determined by the formula:

yI

M

x

xz , (5.2)

N1

N2

N3

N4

N5

N6

N7

I

I

40

40

20

20

30


where Mx – bending moment in the section relative to the x axis; Ix – axial moment of inertia of

the section relative to the x axis; y – ordinate of the point at which the stress is determined.

The value of normal stresses calculated by the theoretical formula and the measured

strains should coincide.

Determining the Displacement of Point A

We determine the displacement of point А using the universal equation of the elastic line

(Koshi–Krylov method). The design diagram of the rod and the diagram of the transverse forces

Qy and bending moments Mx are shown in Figure 5.3.

Figure 5.3 – Design Scheme of the Rod and Diagrams Qy and Mx

The differential equation of the elastic line has the form:

zMIE xx (5.3)

The universal equation of the elastic line is written in the form:

321 3

2

2322

lz

Flz

Fz

FIE x (5.4)

Essentially, there are three equations for sections 1, 2, and 3, nested one into the other.

This recording method leads to the equality of integration constants in all three sections. After

double integration, we have:

3

3

2

3

1

3

126

3

2

26

3

262

lzF

lz

FzFzCCIE x (5.5)

Border conditions:

1. z = 0, υ = 0

F

2

l / 4 y

z

F

2

F

2

F

2 l / 3 l / 3

A

l / 3

F

2

F

2

F

2

F

2 Qy

F·l

2

Mx

31


2. z = l, υ = 0

We substitute condition 1 into the equation for the 1st section, whence С2 = 0.

We substitute condition 2 into the equation for the 3rd section:

333

127

1

27

8

120 lll

FlC (5.6)

18

2

1

lFC

(5.7)

The final equation of the elastic line takes the form:

3

3

2

3

1

32

2

3

2

2

3

236

lzl

zz

zl

EI

F

x

(5.8)

Point A is located on the 1st section, so we substitute the coordinate 4

lz A in the

equation for the 1st section:

xxx

AEI

Fl

EI

Fllll

EI

F 3332

0126,012836

29

42

1

436

(5.9)


1. Turn on the electronic deformation meter a few minutes before the experiment starts.

2. To bring to zero the arrow are presented.

3. Take counts for all seven strain gages at zero load.

4. Turning screw 2 (Figure 5.1), increase the load in steps: 1000N, 2000N, 3000N. The

magnitude of the load is determined by the readings of the indicator connected to the loading

device:

1000N – 0.30 mm

2000N – 0.60 mm

3000N – 0.90 mm.

For each of the three load values, take readings on the scales of the deflection meter and

the electronic deformation meter for each of the seven strain gauges.

5. To process the results of the experiment.

6. Compare theoretical and experimental results and determine the experimental error. To

plot the normal stresses of the beam section.

Note: it is not recommended to take into account the difference in the readings by the

deflection meter when the load changes from 0 to 1000 N, because in this section, the gap in the

system is selected that is not related to the movement of the section under the action of the load.


Table 5.1 – Results of Theoretical Determination of Stresses at the Points of Beam Section I-I

Point № 1 2 3 4 5 6 7

Point y-

coordinate, mm

Stress σ,

MPa

32


Table 5.2 – Experimental Results for Determining Stresses in Section I-I and Displacement of

Section A

Strain Gauge

Number

Force F, N in

ii nK ii E

0 1000 2000 3000

1

n1

Δn1

2

n2

Δn2

3

n3

Δn3

4

n4

Δn4

5

n5

Δn5

6

n6

Δn6

7

n7

Δn7

Deflec-

tion

Meter

nf

fn ffA nKV

Δnf

Control Questions 1. What is called clean and transverse bending?

2. How are the principal stresses determined by the direct transverse bending of the

beam?

3. How are normal stresses theoretically determined in the cross sections of a beam with

direct bending?

4. For what points of the I-beam section of the beam should an additional verification of

strength be carried out for the main and maximum stresses?

5. Describe how the laboratory equipment works.

33


6. OBLIQUE BENDING OF THE ROD


Determination of stress at a point of a rod of rectangular cross-section and complete

displacement of the cross section during oblique bending. Comparison of experimental results

and calculation.


The main element of the laboratory setup is a cantilever-fixed rod (Figure 6.1), loaded

with vertical force.


1 – Rod; 2 – Goniometric Scale; 3 – Strain Gages Group; 4 – Watch Type Indicator (Kf=0.01);

5 – Weights; 6 – Meter of Deformations

x

y φ

А

View А

l

lI

A

x

y

F

I Strain Gages

C

C

34


The design of the support allows you to rotate the rod relative to its longitudinal axis and

fix it in a fixed position. The position of the rod is controlled by an angular scale applied to the

movable part of the rod support.

The angle φ is measured from the vertical. At the free end of the rod, a suspension is

mounted on the cylindrical hinge, on which the loads are stacked when the rod is loaded. The

suspension design allows you to apply force only in the vertical direction.

Four FKPA-1-100 type strain gauges (foil, with a base of 10 mm and a resistance of 100

Ohm) are glued in the longitudinal direction of the rod I (Figure 6.2).

Figure 6.2 – The Strain Gauge Arrangement in Section I

At the free end of the rod, a stand is installed on the base plate of the stand, on which two

dial-type indicators are mounted. Indicators allow you to measure the horizontal and vertical

components of the full movement of point C.

The installation is completed with an electronic deformation meter.


The technical theory of the oblique transverse bending of the rod is based on two

hypotheses: the hypothesis of flat sections and the hypothesis of «not-compressing» the

longitudinal layers on each other in directions perpendicular to them, i.e. on the same hypotheses

as the theory of pure direct bending.

Pure oblique bending of a rod is called one in which the internal forces in the cross

section of the rod are reduced only to the bending moment, the plane of action of which does not

contain any of the main central axes of inertia of the cross section of the rod during bending.

The plane in which the bending moment acts is usually called the force. Unlike direct

bending, the curved axis of the rod does not lie in the force plane, i.e. deformation of the rod

during oblique bending does not occur in the plane of the bending moment, but in the plane

rotated relative to the force by a certain angle towards the plane of least stiffness of the rod

during bending.

Since the hypothesis of flat sections is valid, the cross section rotates about the neutral

axis, remaining flat. The oblique bending line is not perpendicular to the force plane.

Oblique bending is considered as simultaneous bending in two planes zx and zy, where

the x and y axes are the main central axes of inertia of the cross section of the rod. For this, the

bending moment M is decomposed into components with respect to the x and y axes. The normal

x

y N1

N4

A

N3

N2

h

b

35


stress at the cross-sectional point is calculated as the algebraic sum of the stresses due to the

moments Mx and My, i. e.

xI

My

I

M

y

y

x

x , (6.1)

where Mx, My – bending moments relative to the x, y axes; Ix, Iy – axial moments of inertia of the

cross-sectional area of the rod relative to the x, y axes; x, y – the coordinates of the point.

The highest stresses occur at the points of the section farthest from the neutral line.

In oblique bending, the full displacement of a point is defined as 22 uf , (6.2)

where u – the projection of the full displacement of the point on the x axis; υ – the projection of

the full displacement of the point on the y axis.

Consider the oblique bending of the rod.

Bending moment M in section I

IlFM (6.3)

Components of the moment relative to the main central axes of inertia x and y

coscos Ix lFMM (6.4)

sinsin Iy lFMM (6.5)

We define the stress at point А. Since each of the moments Mx and My causes the greatest

tensile stresses at this point, and since h=2b

33

3

1

sin

3

2

cos

b

lF

b

lF

W

M

W

M II

y

y

x

xA

, (6.6)

sin2cos2

33

b

lF IA

(6.7)

At φ = 450, F = 10 N, lI = 650 mm, b = 12 mm, σА = 11,98 MPa.

We define the complete linear displacement of point C.

Consider a cantilever beam of length l, stiffness EIx, loaded at the free end by force Р,

under conditions of direct bending in the zy plane.

To determine the displacement of the point K located at the end of the beam, we use the

differential equation of the elastic line by placing the origin in the embedment.

x

x

EI

zMf , (6.8)

lzPzM x , (6.9)

xEI

lzPf

, (6.10)

We integrate the resulting differential equation

1

2

2Czl

z

EI

Pf

x

, (6.11)

21

23

26CzC

zlz

EI

Pf

x

(6.12)

Border conditions:

1. z = 0, f = 0 ⇒ C2 = 0

2. z = 0, f ' = 0 ⇒ C1 = 0

36


We finally get

2

3

32zl

z

EI

Pf

x

(6.13)

Setting zk = l, we obtain

x

kEI

Plf

3

3

(6.14)

Consider the rod used in this experimental work.

Projections of the complete movement of point C on the x and y axis

y

cEI

Flu

sin

3

1 3

, (6.15)

x

cEI

Fl

cos

3

1 3

(6.16)

Total offset of point C of the rod

2

2

2

2322 cossin

3

1

xy

cccIIE

Fluf

(6.17)

For φ = 450 and Ix = 4Iy

y

cEI

Ff

12

17sin (6.18)


1. To calibrate the strain gauge, it is necessary to set the beam in the straight bend

position (φ=00 or φ=900). It is more advantageous to position the beam so that the bending occurs

in the plane of least rigidity, then the readings will be larger and the accuracy higher.

2. When loading the beam through 10 N from 0 to 40 N, take readings of the strain gauge

for strain gauges located on the upper and lower surfaces of the beam at each load stage.

3. Rotate the beam around the axis to the working position (most often 450 with vertical).

4. Consistently loading the beam with forces of 0, 10 N, 20 N, 30 N, 40 N, to take, at

each load, the readings of the strain gauge for all four strain gages, as well as horizontal and

vertical indicators installed at the end of the beam.

5. To carry out theoretical calculations and processing the results of the experiment.

6. Compare the results of calculation and experiment.

To present the results, it is recommended to use tables 6.1–6.2.

37


Table 6.1 – Deformation Meter Calibration Results

Strain Gauge

Number

Force F, N

in 2

inn

0 10 20 30 40

1 n1

Δn1

2 n2

Δn2

The experimental value of the linear strain per loading stage nK

The theoretical value of linear strain per loading stage E

Normal stress xIxx WlFWM

The scale division of the strain gauge nEK

Table 6.2 – Experimental Stress and Displacement Values

Control Questions

1. What type of loading is called bending?

2. What is the fundamental difference between flat and oblique bending?

3. How to determine the magnitude and direction of full deflection during beam bending?

4. What geometric characteristics are used to determine beam deflection?


Force F, N in

ii nK 0 10 20 30 40

Str

ain G

auge

Num

ber

1 n1

Δn1

2 n2

Δn2

3 n3

Δn3

4 n4

Δn4

Def

lect

ion

Met

er Vert.

nv vn vfv nKf

Δnv

Horiz. nh

hn hfh nKf Δnh

Norm. stress at point А per loading stage

22

4321 EA

Full displacement of point С per loading stage 22

hvC fff

38


7. STRESSES AND DISPLACEMENTS IN A FLAT FRAME


Determination of stresses, displacements and support reactions in a statically

determinable and statically indeterminate flat frame.


The main element of the laboratory setup is a flat frame (Figure 7.1), consisting of three

rectangular cross-section rods rigidly fastened together. The material of the rods is steel 45. The

frame has the following dimensions: l = 300 mm, h = 5 mm, b = 30 mm, a = 150 mm.


1 – Flat Frame; 2 – Watch Type Indicators (Kf=0.01); 3 – Weights;

4 – Loading Mechanism (Ks=0.4 N/mm); 5 – Meter of Deformations (Kε=5·10–7)

The frame is mounted on two articulated supports superimposing three links on the

frame. The design of the support А allows you to impose an additional horizontal connection, i.e.

allows the transition to a statically indefinable flat frame. The kinematic diagram of the support

А is shown in Figure 7.1. Weights fixed on the screw can move when the screw rotates along its

axis. Scale S, mounted parallel to the axis of the screw, indicates the location of the goods

relative to their original position. The moment created by the loads relative to the axis of rotation

l

b

h x

B-B

Strain Gauges

l

a a B

A

l

S

39


of the linkage system is converted by this system into a horizontal force acting on the frame in

the support A. To measure the horizontal movement of the support A, a clock-type deflection

indicator mounted on an arm is used.

In the section In the frame, two strain gauges KFP1-10-100 (foil, base 10 mm, resistance

100 Ohm), which are used to measure deformations in this section of the frame.

Strain gages are connected to an electronic deformation meter.

The frame is loaded with weights of 1 kg.


1. Statically determinable flat frame.

Figure 7.2 – Design Scheme of a Statically Determinable Flat Frame

A flat frame consists of three rods rigidly interconnected, whose geometric axes lie in the

same plane (Figure 7.2). Connections are made in the same plane and external loads act.

Bending moment, transverse and normal forces arise in the cross sections of the frame.

The influence of transverse and normal forces on the strength and stiffness of the frame is

neglected due to smallness and only bending moments are taken into account.

The normal cross-sectional stress is calculated by the formula

yI

M

x

xz (7.2)

We plot the bending moments Mx and calculate the maximum normal stress in the section

B of the frame:

x

x

W

Mmax (7.3)

The moment of resistance of the cross section to bending:

6

2bhWx (7.4)

Then

22max

66

bh

Fa

bh

M x (7.5)

B

F F

a a

l l

l

A

40


We calculate the displacements of sections А and В using the Vereshchagin method. To

do this, we plot the bending moments of the unit load applied at points А and В, respectively, in

the direction of movement. Diagrams of bending moments Мx1A, Мx1

B and Мx are shown in

Figure 7.3.

Figure 7.3 – Bending Moment Diagrams Мx1A, Мx1

B and Мx

l l

1 1

l

l l

l Mx1A

1 l

2

1

2

1

2

Mx1B

F F F·a

a l–a

B

A

F F

Mx

41


Horizontal movement of section A

alEI

Fal

x

A 2 , (7.6)

where 12

3bhI x – axial moment of inertia of the cross section.

Vertical movement of section B

32

22 a

lEI

Fa

x

B (7.7)

2. Statically indefinable flat frame.

Statically indeterminable is such a system, the efforts in all elements of which cannot be

determined only from the equations of statics. The difference between the number of unknowns

(bond reactions and internal force factors) and the number of independent static equations is

called the degree of static indeterminacy.

The number of bonds at which kinematic immutability of the system is achieved is called

the required number of bonds. Any connection imposed in excess of the necessary is called

additional. The number of additional bonds is equal to the degree of static indeterminacy.

To reveal the static indeterminacy of bar systems, the force method is widely used, the

idea of which is that a given statically indeterminate system is freed from additional connections

(both external and internal), and their action is replaced by forces and moments. The values of

these forces and moments are selected so that the displacements correspond to those constraints

that are imposed on the system with dropped connections. Thus, with this method of revealing

static indeterminacy, forces are unknown.

The disclosure of the static indeterminacy of a system begins with discarding additional

relationships. A system freed from external forces and additional connections is called the main

one. It should be statically definable and kinematically unchanged. Then, instead of the dropped

connections, the unknown system needs to apply unknown forces and moments. In those sections

where linear displacements are forbidden, forces are introduced where angular displacements are

forbidden, moments are introduced. Unknown force factors are denoted by Xi, where i – the

number of the unknown. The largest value of i is equal to the degree of static indeterminacy.

When removing internal connections, mutual forces and moments are applied.

The determination of unknown force factors is carried out using the canonical equations

of the force method. The meaning of these equations is the absence of displacements in the

direction of the discarded additional bonds. The number of these equations is equal to the degree

of static indeterminacy of the problem in question.

For n-times a statically indefinable system, canonical equations are written as follows:

0

0

0

2211

22222121

11212111

nFnnnnn

Fnn

Fnn

XXX

XXX

XXX

(7.8)

where δij – displacement in the direction of the i-th force factor from the action of the j-th force

factor equal to 1.

To determine δij, the Mohr integral or, if possible, the Vereshchagin rule is used.

We obtain a statically indefinable flat frame by imposing an additional horizontal

connection in section A (Figure 7.4).

42


Figure 7.4 – Design Scheme of a Statically Indeterminable Flat Frame

The disclosure of the static indeterminacy of a flat frame is carried out by the force

method.

The canonical equation of the force method

01111 FX (7.9)

The selected basic and its equivalent system, as well as diagrams are shown in Figure 7.5.

Figure 7.5 – The Selected Main System (a) and the Corresponding Equivalent System (b)

The coefficients of the canonical equation are determined by the method of Vereshchagin

xEI

l 3

113

8 , (7.10)

alEI

Fal

x

F 21 , (7.11)

2

11

11

2

8

3

l

alFaX F

(7.12)

Plots of bending moments from a given MxF and unit Mx1 нагрузок loads, as well as the

total loads MxΣ of a statically indeterminable flat frame are presented in Figure 7.6.

F F

X1

b) a)

B

F F

a a

l l

l

A

43


Figure 7.6 – Diagram of Bending Moments of a Statically Indeterminable Flat Frame

F F F·a

a l–a

B

A

F F

MxF

l l

1 1

l

l

Mx1

MxΣ

F F

F F

9

32 F

9

32 F

7

32 F·l

9

32 F·l

l–a= l

2

44


The maximum normal stress in the cross section Of the frame is

l

a

bh

Fa

8

3

4

162max (7.13)

We calculate the ratio of maximum stresses in section B for statically definable and

statically indeterminate frames, which characterizes the effect of an additional bond in section A

on the stress-strain state:

l

a

8

3

4

1

1

.st.indetmax

.st.detmax

(7.14)

Note that the introduction of an additional connection significantly reduces the level of

maximum stress and, therefore, makes it possible to create less metal-intensive structures (other

things being equal).

Let us determine the linear vertical movement of the section B of the frame according to

the method of Vereshchagin. The plot of the unit load applied at point B is shown in Figure 7.7.

38

3

42

22 ala

l

EI

Fa

x

B (7.15)

Figure 7.7 – Diagram of Bending Moment from Unit Load at Point B


1. Turn on the electronic deformation meter a few minutes before the start of the

experiment.

2. Set the loads in the initial position (zero on the S scale), for which rotate the screw on

which the loads are fixed, to combine the risks applied to the left load with zero division of the S

scale, which corresponds to the absence of horizontal connection in section A.

3. Check the performance of the deflection meters by tapping lightly on the frame. Set the

scales of both indicators to zero division.

4. Read the strain gauge at zero load.

5. Load the frame sequentially with a force of 10 N and 20 N, taking and recording the

readings of all instruments (deflection meters and strain gauge).

6. Unload the frame. Go on to exploring a statically indefinable frame. Read instrument

readings at zero load.

l

2 1

1

2

1

2

Mx1B

l l

45


7. When loading the frame sequentially with forces of 10 N and 20 N, rotate the screw

with loads, achieving zero movement of section А, which is controlled by the initial position of

the arrow of the deflection meter in section A.

8. At both loads, take readings of devices, including the scale S of the compensating

device, which determines the horizontal force acting on cross-section А.

9. After the experiment is over, completely unload the frame and return the cargo screw

to its original position.

10. To process the results of the experiment in the sequence indicated in the journal.

Perform all theoretical calculations.

11. Compare the results of calculation and experiment, calculate the errors.


Table 7.1 – Statically Determinable Flat Frame

Force

F, N

Weights

Displacement Strain Gauge

Section A

Deflection Meter

Section B

Deflection Meter

S, mm ΔS, mm nε Δnε nA ΔnA nB ΔnB

0

10

20

ΔF=10 S n An Bn

Linear strain on the loading stage nK

Normal stress per load stage E

Section А linear displacement А AfA nKf

Section B linear displacement B BfB nKf

Table 7.2 – Statically Indeterminable Flat Frame

Force

F, N

Weights

Displacement Strain Gauge

Section A

Deflection Meter

Section B

Deflection Meter

S, mm ΔS, mm nε Δnε nA ΔnA nB ΔnB

0

10

20

ΔF=10 S n An Bn

Reaction of removed link to loading stage SKX s1

Linear strain on the loading stage nK

Normal stress per load stage E

Section А linear displacement AfA nKf

Section B linear displacement BfB nKf

Control Questions

1. What is a frame? What types of frames do you know?

2. What frames are called statically determinable and indeterminable?

3. What is the reason for the static indeterminacy of systems?

4. How to determine the degree of static indeterminacy?

5. What is the physical meaning of the canonical equations of the force method?

46


8. STRESSES IN A FLAT GREAT CURVATURE ROD


Determination of stresses in a rod of large curvature during its eccentric tension.


The main element of the laboratory setup (Figure 8.1) is a flat rod of great curvature of

rectangular cross section. Rod material – steel 45. The rod has the following dimensions: R1=40

mm, R2=120 mm, h=80 mm, b=10 mm, l=200 mm.


1 – Great Curvature Rod; 2 – Watch Type Indicator (Kf=0.01); 3 – Screw of the Loading

Mechanism; 4 – Strain Gages Group; 5 – Meter of Deformations (Kε=8.6·10–7)

A A

R1

R2

rn

r0

l

σr

σr

σ

σ

h C

b

А-А

47


The rod is fixedly mounted on the base plate of the laboratory bench.

The loading of the rod is carried out by means of a screw having a right and left thread.

By rotating the screw in one direction or another, eccentric tension or compression of the curved

rod is carried out.

The force F created by the screw is determined using the dial indicator and calibration

table 8.1.

Table 8.1 – Calibration Values of the Loading Mechanism

Watch Type Indicator Values, mm Force F, N

0.21 500

0.63 1500

1.10 2500

An indicator mounted on a curved rod measures the divergence (with eccentric tension)

or the approach (with eccentric compression) of the ends of the curved rod. Thus, the test sample

(curved rod) is a force meter (dynamometer).

KF5P1-10-100 type strain gages (foil, base 10 mm, resistance 100 Ohm), necessary for

measuring deformations and determining normal stresses, are glued to the surface of the rod in

section B-B (Figure 8.2).

Figure 8.2 – The Arrangement of Strain Gages

The compensation strain gage is glued to a special plate that is not deformable during

testing and is located near the working strain gages. Strain gages are connected to an electronic

deformation meter.


Curved rods are divided into rods of small curvature and rods of great curvature. The

main criterion for such a separation is the ratio of the height of the cross section in the plane of

curvature to the radius of curvature of the geometrical axis of the rod r0. If the relation is 5

1

0

0 r

h

, the rod is considered to be a rod of small curvature, and the theory of bending of a straight rod

is used to calculate it. If the relation is 5

1

0

0 r

h, the rod is considered to be a rod of great

curvature, and the bending theory specially developed for rods of this type is used to calculate it.

The theory of bending of a rod of large curvature, studied in the course «Strength of

Materials», is based on the hypothesis of flat sections and the hypothesis of «non-compression

longitudinal layers» on each other.

N2 N3 N4 N5 N6

16 12 12 12 12 16

N1 N7

R1

48


Consider the net bending of the rod in the plane of its curvature. Under the action of

moments М (Figure 8.3), the region of the rod remote from the center of curvature is stretched,

and the region closer to the center of curvature is compressed (the bending moment М is

traditionally considered positive if it increases the curvature of the rod); therefore, there is a layer

that is the boundary between the stretched and compressible regions of the rod. This layer is

known to be called neutral.

Figure 8.3 – Pure Bending of the Rod

We show that with a clean bend of a rod of large curvature, the neutral layer does not

contain the geometric axis of the rod. Assume the opposite - the neutral layer contains the

geometric axis of the rod. Then, according to the hypothesis of flat sections of the diagram, the

changes in the length Δl of the longitudinal layers of the rod as a result of deformation will be

linear. Linear deformation ε of the longitudinal layer of the rod

l

l , (8.1)

where l – the length of the corresponding layer before deformation.

The length of a layer l of a curved rod, in contrast to a straight one, depends on the

distance of this layer to the center of curvature; therefore, the diagram of linear strains has the

form of a hyperbola. Consequently, the diagram of normal stresses E will also be

represented by a hyperbola. The stress state at the point of the curved rod is accepted on the basis

of the hypothesis of non-compression of the longitudinal layers uniaxial. Thus, the accepted

assumption about the location of the neutral layer leads to the fact that the normal force

A

dAN is not equal to zero, which contradicts the loading conditions of the considered rod

(here А – the cross-sectional area of the rod). The condition 0 A

dAN will be met if the

neutral layer is displaced to the center of curvature by a distance e so that the area of the diagram

σ on the left and right of the axis are equal.

Offset value

nrre 0 (8.2)

where rn – radius of curvature of the neutral layer, which is determined by the formula

A

n

r

dA

Ar (8.3)

M M

r0

rn

R1

R2

Δl + – ε σ + – + –

C

C1 e

A1

A2 F2

F1

F1 – Resultant Tensile Force

F2 – Resultant Compression

Force

A=A1+A2 – Section Area

49


where hbA ; r – distance from the point of the rod to the center of its curvature; for a

rectangular cross-section drbdA .

Thus,

1

2lnR

R

hrn

(8.4)

Since the value of e is determined by the difference between two close (especially in the

case of a relatively small curvature of the rod) quantities r0 and rn, it is necessary to know their

value with high accuracy (4–5 significant digits).

The normal stress σ at the cross-section points of the rod at a distance r from the center of

curvature is determined by the formula

r

rr

eA

M n

(8.5)

When loading the test rod (Figure 8.4) in cross-section В-В, a bending moment М и and

normal force N arise.

Figure 8.4 – The Design Scheme of the Rod

From equilibrium conditions

FN (8.6)

lrFM 0 (8.7)

The normal stress σ in the cross-section В-В is

A

F

r

rr

eA

lrF n

0 (8.8)

The solution to the problem of bending a rod of large curvature of a rectangular cross-

section by methods of elasticity theory, taking into account the interaction of longitudinal layers,

was obtained by Kh. F. Golovin in 1881.

According to this solution, a biaxial stress state occurs in the rod, characterized by

stresses σr and σt:

r

RR

R

rR

R

R

r

RR

Cb

Mr

12

1

2

2

2

1

2

2

2

2

2

1 lnlnln4

, (8.9)

2

1

2

212

1

2

2

2

1

2

2

2

2

2

1 lnlnln4

RRr

RR

R

rR

R

R

r

RR

Cb

Mt , (8.10)

where 2

1

22

2

2

1

22

1

2

2 ln4

R

RRRRRC

l

F

F

r0

r0

N=F

Mx=F(l+r0)

F

50



1. Experimental work.

Connect the electronic deformation meter to the strain gauges (upper connector). Connect

an electronic deformation meter to the electric current network.

Check the performance of the strain gauge by loading a curved beam with a screw of

arbitrary (small) force and observing the change in the instrument readings. Load the beam with

a force of 0.5 kN (for eccentric tension, turn the screw counterclockwise, for eccentric

compression clockwise). Read the strain gauge for seven strain gauges and record these readings

in the laboratory journal.

Load the beam sequentially up to 1.5 kN and 2.5 kN and take readings of the strain gauge

at each force value, recording them in the laboratory log.

2. Processing the experimental results.

Calculate the difference in readings Δni for each strain gauge corresponding to the

loading stage ΔF = 1 kN.

Calculate the arithmetic mean of the difference in .

Determine the relative strain increment Δεi for each strain gauge corresponding to the

load stage ΔF = 1 kN using the formula Δεi = Kε · Δni, where Kε – scale division price of the

electronic deformation meter.

Calculate the normal stress increment Δσi at each point of the beam in which the strain

gauge is glued, using the relation

A

F

r

rr

eA

lrF

i

nii

0 (8.11)

(eccentric extension).

Enter the results in a table.

3. Comparison of experimental and theoretical results.

To compare experimental and theoretical results, it is necessary to plot a diagram of

normal stress increments (Figure 8.5) based on a theoretical calculation. Then, using the

experimental values of the normal stress increment, plot the experimental points on the graph.

They should be located near the theoretical curve.

Figure 8.5 – Normal Stress Diagram

N1

N2 N3 N4 N5 N6

N7

40

30

20

10

0

–10

–20

Theoretical

Curve

Experimental

Values

51


Calculate the relative errors of normal stresses at points 1 and 7 according to the relation

%100exp

theorexp

(8.12)

To present the results of the experiment, it is recommended to use tables 8.1– 8.2.

Table 8.1 – The Results of the Theoretical Calculation of the Stress at F=1000 N

Parameter № of Point (Strain Gauge Number)

1 2 3 4 5 6 7

Coordinates of point r, mm

0AFN , MPa

r

rr

eA

lrF nM

0

0 , MPa

MN , MPa

Table 8.2 – Experiment Results

Strain Gauge

Number

Force F, N

in ii nK ii E

500 1500 2500

1 n1

Δn1

2 n2

Δn2

3 n3

Δn3

4 n4

Δn4

5 n5

Δn5

6 n6

Δn6

7 n7

Δn7

Control Questions

1. By what criteria can a rod be considered a rod of great curvature?

2. What is the theory of bending of a rod of large curvature based on?

3. What is a neutral layer?

4. How is the normal stress determined at the points of the cross section of a rod of large

curvature?


52


9. STRESS STATE WITH JOINT BENDING AND TORSION OF A ROD


Determination of the values of the principal stresses and the position of the principal axes

at a point on the surface of the rod during its joint bending and torsion by the method of

electrotensometry.


The main element of the laboratory setup (Figure 9.1) is a tubular rod, loaded with

bending and torque.


1 – Tubular Rod; 2 – Support Screw-Clamp; 3 – Lever;

4 – Strain Gages Group; 5 – Weights; 6 – Meter of Deformations

y

x

D

h

Sec. I

Sec. I

lI

a

A

53


At point А of section I, three strain gages are glued to measure strain (Figure 9.2).

Figure 9.2 – The Strain Gauge Arrangement at Point A of Section I

Deformation is measured using an electronic deformation meter. To calibrate the scale of

the electronic deformation meter, the installation design provides for the possibility of installing

a hinged support on the free edge of the rod, in which case the rod is loaded only with torque.

Equipment geometrical parameters: D=57 mm, h=1 mm, lI=300 mm, a=450 mm.

Correspondingly, the geometric characteristics of the cross section are Wx=2550 mm3,

Wt=5100 mm3. Poisson's ratio for the rod material v=0.33.


1. Determination of the magnitude and direction of the main stresses by the method of

electrotensometry.

The method of electrotensometry is based on the measurement of strains using strain

gauges. The sensitive element of each strain gauge is a conductor, which is tightly glued to the

element under study and is deformed with it. A change in the geometric parameters of the

conductor leads to a change in its electrical resistance, which changes the current in the circuit

into which the conductor is included. This change in current is proportional to strain and is

recorded by a deformation meter. Typically, the inclusion of a strain gauge in the device is

carried out according to a bridge circuit using a compensation strain gauge, which eliminates the

influence of external temperature, which can significantly change the electrical resistance (this

does not matter with the compensation circuit, since both strain gauges operate at the same

temperature).

At any point on the surface of the body, loaded arbitrarily, a plane stress state occurs (in

the absence of force on the surface of the body). In this case, one main site and the main stress is

known: 0I

main , and the magnitudes and directions of two other main stresses must be found;

Thus, the task is to determine three quantities: two values of the main stresses and the angle

between the original and main axes.

Three equations are needed to solve it, therefore, it is necessary to measure the strains in

the direction of the three axes u, v, z passing through a given point А. These deformations are

measured using three strain gauges glued at the test point (Figure 9.2). It should be noted that the

stress and strain state in the general case of loading varies from point to point, therefore, the

strain gauge fixes the average strain at points corresponding to its base. To avoid significant

errors, a strain gauge should be chosen so that its base is substantially smaller than the strain

gradient. In our case, this is achieved by the fact that the base of the strain gauge is small in

comparison with the geometric dimensions of the object under study.

z

x

A

σ1

σ3 σ1

σ3

450

450

N1

N2

N3

1 3

54


Deformation in an arbitrary direction is associated with deformations in the direction of

the main axes by a simple geometric relationship (Figure 9.3).

Figure 9.3 – The Geometric Ratio of Deformations in an Arbitrary Direction and the Direction of

the Main Axes

22 sincos IIIIIs , (9.1)

where α – the angle between the axes II and s.

After a simple trigonometric transformation we get

2cos22

IIIIIIIIIIs

(9.2)

For the u, v, z directions where deformations are measured, this ratio is converted to the

following:

2cos22

IIIIIIIIIIu

, (9.3)

0452cos22

IIIIIIIIIIz

, (9.4)

0902cos22

IIIIIIIIIIv

(9.5)

The solution of the system of equations (9.3-9.5) with respect to εII, εIII and α gives

22,

2

1

2zvzu

vuIIIII

, (9.6)

uv

vuztg

22 (9.7)

Then the values of the main stresses are found using the generalized Hooke's law:

IIIIIII vv

E

21, (9.8)

IIIIIIII vv

E

21 (9.9)

Then we assign indices 1, 2, 3 to the main stresses so that σ1≥σ2≥σ3.

III

II 0

B1

B

A A1

C

C'

C1

S

α

55


2. Analysis of the stress state in a thin-walled pipe during its bending and torsion.

The stress state at point A, located on the surface of the test rod, is shown in figure 9.4.

Figure 9.4 – Stress State at Point А

Normal and tangent stresses are determined by the formulas:

x

xz

W

M max , (9.10)

P

tzx

W

M max (9.11)

where Ix lFM , aFM t – bending and torsional moments in the cross-section I,

4

2hDWx

– moment of cross section resistance to bending, 2

2hDWP

– torsion

resistance of the cross section.

The main stresses and the angle of inclination of the main axis to the longitudinal axis of

the rod are calculated by the formulas following from the Mohr theory for the main stresses in

the areas perpendicular to one of the main ones (in our case, this is a stress-free surface, σ2 = 0):

22

3,122

, (9.12)

(the structure of this formula immediately shows that one of the stresses determined by it is

positive, and the other is negative, which, in combination with the already known third main

stress, equal to zero, gives reason to assign indices 1 and 3 to them),

22 tg , (9.13)

045 (9.14)

The last relation is due to the fact that the angles α and β are counted from the u and z

axes (in different directions), the angle between which is 450.

y

x

z

τzx

σz

τxz

56



1. The determination of the division price of the electronic deformation meter.

The equipment is brought into a state corresponding to loading by one torque (a thrust

bearing is installed that prohibits vertical movement of the free end of the rod – Figure 9.5).

Figure 9.5 – The Design Scheme of the Rod During Graduation

(stress state at point A – net shift)

After that, loading is carried out sequentially by forces of 50, 100, 150, 200 N. At each

load, readings of strain gauges №1 and №3 (located along the axes u and v) are taken. Since a

pure shear stress state is realized under such loading, the average difference in readings

corresponding to a load of 50 N should be approximately the same in modulus, and the

arithmetic mean of the differences in the reports on sensors 1 and 3 taken modulo is taken as the

average Δn. The division price is determined by the formula

nE

vK

1 , (9.15)

where Δσ=Δτ – average stress increment at point А in the direction of the axes u and v

corresponding to a load of 50 N, Δn – average difference in readings from the deformation

meter, E and ν – elastic constants of the material.

2. Experiment and processing of its results.

After removing the thrust bearing, load the system with forces of 50, 100, 150, 200 N

(Figure 9.6). At each load, take the readings of the deformation meter on each of the three strain

gauges (along the axes u, z and v).

Figure 9.6 – The Design Scheme of the Rod and the Diagram of Internal Force Factors

Sec. I

lI

a

F

y

x A Mx

F·lI

l

F·l Mt

F·a

Sec. I

lI

a

F

y

x A y

x

Sec. I A

Mк τ

Mt

y

x z

τzx=τ τxz=τzx=τ

Point А

F·a

Free Surface

57


Calculate strain values

uu nK , (9.16)

zz nK , (9.17)

vv nK (9.18)

Find the principal strains ε1 and ε3 by the corresponding formulas:

E

vv

E

113211

(9.19)

E

vv

E

112133

(9.20)

Determine the angle β between the source and main axes.

Determine the main stresses by the formulas of the generalized Hooke's law.

3. Theoretical Results.

Calculate the principal stresses and angle β using the theoretical calculation formulas.

4. Comparison of theoretical and experimental results.

Combine the data of the theoretical calculation and experiment into a table and calculate

the relative errors for the two principal stresses and angle β.


Table 9.1 – The Results of Determining the Scale Division of an Electronic Deformation Meter

Force F, N Strain Gauge №1 (axis 1) Strain Gauge №3 (axis 3)

n1 Δn1 n3 Δn3

0

50

100

150

200

ΔF=50 N 1n 3n

The scale division of the electronic deformation meter nK

Table 9.2 – The Results of Determining the Stress-Strain State at the Point of the Rod During

Bending and Torsion

Force F, N Strain Gauge 1 (axis u) Strain Gauge 2 (axis z) Strain Gauge 3 (axis v)

nu Δnu nz Δnz nv Δnv

0

50

100

150

200

ΔF=50 N un zn vn

Main stresses 2

311 1 vvE 2

133 1 vvE

Control Questions 1. What is complex stress state?

2. What are the main stresses?

3. What are major axes?

4. What stresses act in the longitudinal sections of the rod under the simultaneous action

of torque and bending moments?


58


10. EXPERIMENTAL VERIFICATION OF THE RECIPROCITY THEOREM


An experimental verification of the validity of the reciprocity theorem by the example of

displacements in a flat frame under its different loads.


The main element of the laboratory setup is a flat frame (Figure 10.1), consisting of three

rectangular cross-section rods rigidly interconnected.


1 – Flat Frame; 2 – Watch Type Indicators (Kf = 0.01); 3 – Weights;

4 – Loading Mechanism (Ks = 0.02 N·m/mm)

l l b

h x

l

B

A

S H

59


The frame is mounted on two hinged supports. The design of the support A allows you to

apply a load in section A in the form of a concentrated moment. This is achieved by shifting the

weights secured to the screw as it rotates. The scale S mounted on the support A parallel to the

axis of the screw indicates the location of the goods relative to their initial position at which the

moment is zero.

Frame parameters: l = 300 mm, h = 5.1 mm, b = 30 mm, H = 98 mm. Material: steel St3,

elastic modulus Е = 2·105 MPa

To measure the deflection in section B and the angle of rotation of section A, watch type

indicators are used.

The frame is loaded with weights of 1 kg.


The reciprocity theorem was proved by the Italian scientist E. Betty in 1872, therefore it

is often called the Betty theorem. According to the reciprocity theorem, the work of the forces of

the first state on the corresponding displacements of the second state is equal to the work of the

forces of the second state on the corresponding displacements of the first state.

In accordance with the statement of the theorem, we can write

12 2211 FF FF , (10.1)

where F1 – system of forces of the first state, F2 – system of forces of the second state, δ1F2 –

displacements in the second state corresponding to the forces of the first state, δ2F1 –

displacements in the first state corresponding to the forces of the second state.

Betty's theorem is valid for linear systems in which the displacements are small compared

to the overall dimensions, and the Hooke law is valid for the material of the system.

A rod system is used as a linear system – a flat frame located in series in two states. The

first state is characterized by the loading of the frame by the force F applied at point B (Figure

10.2).

Figure 10.2 – Design Scheme of the Frame and the Diagram of Bending Moments in State «1»

F B

l l

A

F Fl

2

Mx

F

2

F

2

A

B

1

1

2l

1

2l

1

2

1 1

1

1

Mx1

l l

60


The second state is characterized by the loading of the frame by the moment M applied at

point A (Figure 10.3).

Figure 10.3 – Design Scheme of the Frame and the Diagram of Bending Moments in State «2»

In the first state, section А is rotated through an angle θA1, which corresponds to the

moment М in the second state. In the second state, section В receives a displacement fB2, which

corresponds to the force F in the first state.

The work of the force F on the displacement fB2 is

221 BfFA (10.2)

The work of the moment М on the displacement θA1 is

112 AMA (10.3)

According to the reciprocity theorem

12 AB MfF (10.4)

The theoretical displacements fB2 and θA1 are found by the Mohr method (Vereshchagin

rule).


1. Set movable loads in the initial position (zero on the S scale), for which it is necessary

to rotate the screw on which the loads are fixed in one direction or another until the risks applied

to the left load are combined with zero division of the S scale. This corresponds to the absence of

moment M in the cross section A.

2. Check the operability of the deflection meter and the goniometer, for which lightly tap

the frame and observe the movement of the large hand of the dial indicator. The arrow should

fluctuate around one arbitrary division of the scale.

3. After making sure that the devices are working, take their readings and write these

readings in the laboratory journal.

4. Load the frame with the force F = 20N (state «1») and take readings of the goniometer

indicator (section A).

5. Unload the frame. The experiment is repeated two more times.

1 B

l l

A

F l

2

Mx2

1

2

1

2

A

B

M

M

2·l

M/2

M M

M

M

Mx l

M

2·l

61


6. Load the frame with moment M (state «2»), for which it is necessary to rotate the

screw with moving loads until the risks on the left load are combined with the division of the S

scale equal to 100 mm. Read the deflection indicator.

7. Unload the frame by returning the moving loads to their original position. Repeat the

experiment two more times.

8. Calculate the arithmetic mean values of the differences in the readings of the deflection

meter, goniometer and S scale corresponding to the loading stage, and write these readings in the

laboratory journal.

9. Determine the increment of the angle of rotation of section A, the linear displacement

of section B, the work of external force and external moment corresponding to the loading stage.

10. Write the results of the calculation and the experiment in a table, calculate the errors

and draw a conclusion about the validity of the reciprocity theorem (Betty's theorem).


Table 10.1 – Experiment Results (State «1»)

Exp. № Force F, N Goniometer, section A

np Δnp nA1 ΔnA1

1 0 –

20 20

2 0 –

20 20

3 0 –

20 20

F 20 Н 1An

Section rotation angle А HnK AfA 11

Moment work M at displacement 1A 112 AMA

Table 10.2 – Experiment Results (State «2»)

Exp. № Force F, N Goniometer, section B

S, mm ΔS, mm nB2 ΔnB2

1 0 –

100 100

2 0 –

100 100

3 0 –

100 100

S 100 mm 2Bn

External moment SKM S

Section linear displacement B 22 BfB nKf

Force work F at displacement 2Bf 221 BfFA

Control Questions 1. Formulate a theorem on the reciprocity of work and displacement.

2. What is a linear rod system?

3. What is the difference between the two loading states of a flat frame in this laboratory

work?

4. How are the theoretical displacements determined in this laboratory work?


62


11. CRITICAL LOAD FOR A FLEXIBLE COMPRESSED ROD


Verify experimentally the validity of the Euler formula for the critical force during

compression of a flexible rod.


The main element of the laboratory setup is a rod of rectangular cross section. The core

material is steel 45. The dimensions of the rod: l=500 mm, b=35 mm, h=2 mm (Figure 11.1).

The rod is installed in the right and left supports, the design of which allows for hinged

or rigid fastening of the ends of the rod in the plane of least rigidity.


1 – Rod; 2 – Screw of the Loading Mechanism; 3 – Watch Type Indicators (Kf=0.01); 4 –

Adjustment Screws; 5 – Dynamometer (Kd=520 N)

The right support is connected to a dynamometer using a calibrated dial gauge, the left –

to a screw type loading device. In the middle section of the rod there is a deflection meter (dial

type indicator on the counter).

To determine the force F acting on the rod, one needs to use the empirical dependence

obtained when calibrating the dynamometer. For each laboratory installation, this dependence is

individual.


The design scheme of a ideal rod

For a flexible rod, the critical force Fcr is determined by the Euler formula

2min

2

crl

IEF

(11.1)

Three anchoring schemes are considered (figure 11.2).

l

b

h

x

y

63


Figure 11.2 – Schematic of Attachment of a Rod

The design scheme of a non-ideal rod

The real rod is not ideal, because contains imperfections. Two imperfections are of the

greatest practical importance: initial curvature and eccentric application of force.

In a laboratory setup, measures have been taken to reduce the eccentricity of the

application of force to an acceptable level. The eccentricity is 0.1–0.15 mm, therefore, the initial

curvature is of greatest importance for the laboratory setup, because It’s almost impossible to

make a perfectly straight flexible rod. We obtain an approximate formula that allows us to take

into account the influence of the initial curvature of the rod on the dependence «longitudinal

force – characteristic displacement».

Let the rod hinged at the ends have an initial curvature (Figure 11.3).

Figure 11.3 – Initial Curvature of the Rod before Loading

We assume that the geometrical axis of the bar before loading is described by the

function

l

zfu

sin00 (11.2)

The elastic line of the rod after loading:

x

z

f0

l

l/2

u0

z

F Scheme 3

F Scheme 2

F Scheme 1

64


Figure 11.4 – Curvature of a Rod Under Loading by Longitudinal Force

The differential equation of the elastic line under loading by the longitudinal force F has

the form

zMuEI yy , (11.3)

uuFzM y 0 , (11.4)

uuFuEI y 0 , (11.5)

0

22 ukuku , (11.6)

where yEI

Fk 2

.

We write the solution of the inhomogeneous differential equation as the sum of the

general solution of the corresponding homogeneous equation and the particular solution of the

inhomogeneous one.

*21 cossin ukzCkzCu (11.7)

A particular solution of the differential equation can be represented as

l

zBu

sin* (11.8)

Differentiate u* with respect to z

l

z

lBu

cos* , (11.9)

l

z

lBu

sin

2

2

* (11.10)

Hence

l

zfk

l

zBk

l

zB

l

sinsinsin 0

22

2

2

, (11.11)

0

22

2

2

fkkl

B

(11.12)

After transformation

Fl

EI

fB

y

2

2

0

1

(11.13)

If 2

2

Eulerl

EIF

y , then

F

F

fB

Э

1

0 .

x

z

f

l

l/2

u0

z

u f0 F

65


Border conditions

00,0 2 Cuz

00, 1 Culz

We finally get

l

z

F

F

fu

Euler

sin

1

0

(11.14)

Define the cross section deflection at 2

lz .

F

F

ff

Э

1

0 (11.15)

We transform the resulting dependence to the form

EulerEuler F

f

F

f

F

f 0 (11.16)

We obtained a linear dependence F

f on f with an angular coefficient

EulerF

1; the same

dependence fF

f is constructed from experimental data, approximated by a straight line, and its

correspondence to the theory is estimated, which is an experimental test of Euler's formula.

It should be noted that the Euler force FEuler defined by this method coincides with the Fcr

of the ideal rod, under two conditions:

1. The initial curvature of the axis of the rod should lie in the plane of least rigidity.

2. Rod flexibility in the plane of least rigidity

A

I

l

min

(11.17)

must be higher than the limit determined by the proportionality limit of the material σpr,

pr

2

pr

E (11.18)


1. Set up the laboratory equipment.

2. Set the beam deflection indicator to 0.

3. Load the beam with longitudinal force to a deflection value of 1 mm and read off the

dynamometer indicator.

4. Load the beam sequentially with compressive longitudinal force corresponding to

deflections from 1 mm to 3 mm every 0.2 mm, taking readings of the dynamometer indicator

corresponding to each load with an accuracy of 0.001 mm.

5. Slowly unload the rod from compressive forces, while the rod should return to its

original state.

6. Determine the longitudinal force from the readings of the dynamometer indicator using

an individual empirical relationship for a given installation.

7. Build a graph of the experimental results in coordinates (f/F, f).

66


8. Graphically approximate the resulting dependence by a linear function, determine its

angular coefficient, and from it – FEuler.

9. Compare with each other the obtained value of FEuler with the theoretical.

10. Change the way of fixing the beam and repeat the experiment.

It is recommended to use table 11.1 to present the results of the experiment.


f, mm 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

Sch

eme

№1

fF,

mm

F, N

f/F,

mm/N

Sch

eme

№2

fF,

mm

F, N

f/F,

mm/N

Sch

eme

№3

fF,

mm

F, N

f/F,

mm/N

Control Questions

1. What is the difference between ideal and non-ideal rods?

2. What is the elastic line of the rod?

3. What is an experimental test of the Euler formula?

4. What is the plane of least rigidity?


67


12. IMPACT STRENGTH OF MATERIAL


The acquisition of practical skills in conducting tests and calculating the impact strength

parameters of the material.


To test the materials for impact strength, a pendulum driver with a marginal energy

reserve of 30 kgm is used. The speed of the knife of the pendulum should be in the range from 4

to 7 m/s, which corresponds to the rise of the knife in the pendulum to a height of 0.8 to 2.6 m.

The testing machine consists of a cast iron bed with one vertical stand. A heavy

pendulum is suspended on the horizontal axis in the upper part of the rack (Figure 12.1).


1 – Test Sample; 2 – Scale of the Protractor; 3 – Fixing Mechanism;

4 – Pendulum; 5 – Protective Fence

Parameters of the pendulum: weight P = 20 kg, length L = 0.8 m.

At the beginning of the test, the pendulum rises mechanically to its upper initial position

and is held in this position by a special locking mechanism.

The sample is placed horizontally on two steel supports located at the bottom of the

machine rack. The pendulum has a scale by which the angles of the pendulum rise before and

after the fracture of the sample are determined. The working area of the testing machine has a

protective guard and a pan to dump a broken sample.

68


It was established by experiments that the presence of a sharp transition in the product

sections, as well as the first traces of cracking, facilitates the manifestation of brittleness and

promotes premature destruction of the material.

For comparability of the obtained results, the shape and dimensions of the samples,

incisions, and the conditions of their machining are standardized. Requirements for cutting

blanks for samples, their number, position (samples longitudinal, transverse, tangential, radial)

are determined by the standards for the manufacture of products or technical specifications for

materials. Samples cut out of fiber fibers are less sensitive to changes in shape and size than

samples cut along fibers. Different types of test specimens are specified by the standard.

At high loading speeds, for example, upon impact, the risk of brittle fracture increases.

This danger increases with the presence of holes, cracks, grooves and other stress concentrators

in the structure. Therefore, notched specimens are used (Figure 12.2). The notch allows you to

focus all the fracture strain in one place, and also significantly weakens the cross section and

causes an increase in stresses.

Figure 12.2 – Sketches of Test Sample with Various Notches

Incision: U-type (a); Y-type (fatigue crack) (b); V-type (c)

To obtain reliable test results, it is necessary to carefully check the quality of the notch

and to accurately control its cuts with a caliper with a cone having a division price of 0.05 mm.

a)

H1 H

l/2

R

A

A

A-A

B

b)

h H

l/2

A

A

A-A

B

h1

h

h1

450

c)

H1 H

l/2

A

A

A-A

B

450

69


Even slight defects on the notch surface or slight deviations from the standardized

dimensions can have a significant impact on the final test results.

The following factors also influence the test results: the geometric dimensions of the

samples, the distance between the supports, the angle of sharpening and thickness of the knife,

the speed of the hammer, the temperature of the test sample.

When testing the impact strength of notched specimens, the similarity law is completely

inapplicable, since there is no proportional relationship between the cross-sectional area of the

specimen and the amount of work spent on its brittle fracture. For geometrically similar samples,

the impact strength increases with increasing sample sizes. With an increase in the width of the

specimen at the same other sizes, the influence of the notch increases and the toughness

decreases.

An increase in the length of the specimen with a decrease in its height from the notch side

has little effect on the value of impact strength. Of great importance is the observing installation

of the sample on the supports, it is necessary that the top of the notch is exactly in the same plane

as the blade of the hammer knife.

For the correct location of the sample, a metal template is used, which, before testing,

checks the symmetry of the installation of the sample on the supports. Plastic deformation of an

incised specimen during shock bending is concentrated only near the incision in the middle part

of the specimen along its length, therefore, the region of propagation of the deformation largely

depends on the shape and depth of the incisions.

In this laboratory work, samples with a V-shaped notch are tested (Figure 12.3) made of 4

different materials: aluminum, steel, brass and copper. The samples have the following

dimensions: length L = 55 mm, height H=10 mm, width B=10 mm, working section height H1=8

mm.

Figure 12.3 – The Geometric Model of the Test Sample


In the parts of machines and mechanisms exposed to rapidly occurring shocks during

operation, stresses change strongly. Therefore, preliminary tests of the samples of metals and

alloys from which it are made, carried out under the influence of slowly and gradually increasing

static loads, do not make it possible to judge the reliability of the structure as a whole. In

particular, this applies to parts made of such metal grades, which under the influence of certain

service conditions tend to become brittle under the influence of lowering the temperature, the

presence of stress concentrators, increasing the absolute dimensions, increasing the strain rate

and other factors. The insignificant resistance of such parts to high-speed dynamic loads will be

70


especially dangerous for them. Therefore, in order to check in advance the tendency of materials

to brittle fracture, they are subjected to tests under shock loads.


1. Before starting work, familiarize yourself with the design of the machine for testing.

Care must be taken when working on the test machine to avoid accident due to the fall of the

pendulum. Record the type of the pendulum, its weight P, length L, angle of incidence before the

test α in degrees, maximum energy of the machine Аmax. Measure the dimensions of the test

sample.

2. Place the specimen on the supports, while the notch should be located symmetrically

with respect to the supports and face in the direction opposite to the impact.

3. Release the pendulum that falls and destroys the sample. On the scale, fix the angle

formed between the pendulum arrow and the stop equal to the take-off angle or directly on the

pendulum scale, if it is graduated in units of work.

coscos LPA , (12.1)

where P – weight of the pendulum, L – length of the pendulum, α – angle of the pendulum

before the impact, β – angle of the pendulum after the impact.

4. Calculate the specific impact strength of the sample material by the formula:

S

Ad n , (12.2)

where А – impact work spent on breaking a specimen; S – the cross-sectional area of the

specimen (at the incision).

To present the results of the experiment, it is recommended to use table 12.1.


Material

Geometric Characteristics of the Sample Specific Impact

Strength dn,

kJ/m2 B, m H1, m S, m2

Aluminum

Steel

Brass

Copper

Control Questions

1. What is the essence of the method of testing the material for impact strength?

2. What properties of the material characterize large and small values of impact strength?

3. How is impact strength used?

4. What caused the difference in material properties during static and dynamic tests?

5. What is an incision made on a specimen when determining impact strength?

Strength of Materials Virtual Lab

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