Mechanics of Materials 13-1 - Valparaiso University

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13-1Mechanics of MaterialsStress-Strain Curve for Mild Steel


13-2aMechanics of MaterialsDefinitions

• Hooke’s Law

• Shear Modulus:

• Stress:

• Strain:

• Poisson’s Ratio:

• Normal stress or strain =

• Shear stress = || to the surface

!

" to the surface


13-2bMechanics of MaterialsDefinitions

Uniaxial Load and Deformation

Thermal Deformation


13-3aMechanics of MaterialsStress and Strain

Thin-Walled Tanks

Hoop Stress:

Axial Stress:


13-3bMechanics of MaterialsStress and Strain

Transformation of Axes


13-3cMechanics of MaterialsStress and Strain

Five simplified steps to construct Mohr’s circle1. Determine the applied stresses (σx, σy, τxy).2. Draw a set of σ-τ axes.3. Locate the center:

4. Find the radius (or τmax):

5. Draw Mohr’s circle.

!

"c

= 1

2("

x+"

y).


13-3d1Mechanics of MaterialsStress and Strain

For examples 1 and 2, use the following illustration.

Example 1 (FEIM)The principal stresses (σ2, σ1) are most nearly(A) –62 400 kPa and 14 400 kPa(B) 84 000 kPa and 28 000 kPa(C) 70 000 kPa and 14 000 kPa(D) 112 000 kPa and –28 000 kPa



The center of Mohr’s circle is at

!

"max

= (30000 kPa)2+ (24000 kPa)2

= 38419 kPa

!

"1

= "c# $

max= (#24000 kPa # 38419 kPa) = #62419 kPa

!

"2

= "c

+ #max

= ($24000 kPa + 38419 kPa) = 14418 kPa

Therefore, (D) is correct.

Using the Pythagorean theorem, the radius of Mohr’s circle (τmax) is:

!

"c

= 1

2("

x+"

y) = 1

2(#48000 kPa + 0) = #24000 kPa



Example 2 (FEIM):The maximum shear stress is most nearly

(A) 24 000 kPa(B) 33 500 kPa(C) 38 400 kPa(D) 218 000 kPa

Therefore, (C) is correct.

In the previous example problem, the radius of Mohr’s circle (τmax) was

!

"max

= (30000 kPa)2+ (24000 kPa)2

!

= 38419 kPa (38400 kPa)


13-3eMechanics of MaterialsStress and Strain

General Strain

!

Note that "x is no longer proportional to #

x.


13-3fMechanics of MaterialsStress and Strain

Static Loading Failure Theory

Maximum Normal Stress: A material fails if

•

Or

•

!

" # St

!

" # Sc

This is true of brittle materials.For ductile materials:

Maximum Shear

Distortion Energy (von Mises Stress)

!

Tmax

= max"

1# "

2

2,"

1# "

3

2,"

2# "

3

2

$

%

& &

'

(

) )

>S

yt

2

!

" # = 1

2#

1$ #

2( )2

+ #1$ #

3( )2

+ #2$ #

3( )2%

& ' (

) * > S

yt


13-3gMechanics of MaterialsStress and StrainTorsion

• For a body with radius r beingstrained to an angle φ, the shearstrain and stress are:

!

" = rd#

dz

!

" = G# = Grd$

dz

• For a body with polar moment ofinertia (J), the torque (T) is:

!

T = Gd"

dzr

2

dAA# = GJ

d"

dz

The shear stress is:

!

"#z

= GrT

GJ=

Tr

J

• For a body, the general angulardisplacement (φ) is:

!

" =T

GJdz

0

L

#

• For a shaft of length (L), the totalangular displacement (φ) is:

Torsional stiffness:


13-3hMechanics of MaterialsStress and Strain

Hollow, Thin-Walled Shafts


13-4aMechanics of MaterialsBeams


13-4bMechanics of MaterialsBeams

Load, Shear, and Moment Relations

Load:

Shear:

For a beam deflected to a radius of curvature (ρ), the axial strain at adistance (y) from the neutral axis is

!

"x

= #y /$.


13-4c1Mechanics of MaterialsBeams

Shear and Bending Moment Diagrams

Example 1 (FEIM):Draw the shear and bending moment diagrams for the following beam.



Shear is undefined at concentrated force points, but just short of x = 12 m

!

Rl+ R

r= 100

N

m

"

# $

%

& ' 16 m( ) = 1600 N

Rl= (8)(R

r(4) = 0

Therefore, Rl = 533.3 N and Rr = 1066.7 N

So the shear diagram is:

!

From 0 m to 12 m, V = Rl" 100

N

m

#

$ %

&

' ( x = 533.3 N" 100

N

m

#

$ %

&

' ( x; 0 m < x < 12 m

!

V (12") = 533.3 N" 100N

m

#

$ %

&

' ( (12 m) = "666.7 N

!

V = 1600 N" 100N

m

#

$ %

&

' ( x; 12 < x ) 16 m

!

From 12 m to 16 m, V = V (12") + Rr" (100 N)(x "12)



The bending moment is the integral of the shear.

!

M = Sdx0

12

" + Sdx12

x

" = #800 + 1600 N#100N

mx

$

% &

'

( )

0

12

" dx

!

= "800 N #m + 1600 N( )x " 50N

m

$

% &

'

( ) x

2$

% &

'

( )

12

x

M = 533.3x – 50x2; 0 m < x < 12 m

!

M = "800 N #m + 1600 N( )x " 50N

m

$

% &

'

( ) x

2 " 1600 N( ) 12 m( ) + 50N

m

$

% &

'

( ) 12 m( )

2

!

M = "12800 N #m + 1600 N( )x " 50N

m

$

% &

'

( ) x

2

!

12 m < x " 16 m

Or, let the right end of the beam be x = 0 m

!

Then, S = " 100N

m

#

$ %

&

' ( x; " 4 m < x ) 0 m

!

M = Sdxx

0

" = # 100N

m

$

% &

'

( )

x

0

" x = # 50N

m

$

% &

'

( ) x

2



The bending moment diagram is:


13-4d1Mechanics of MaterialsBeams

Example 2 (FEIM):The vertical shear for the section at the midpoint of the beam shown is(A) 0(B) (C) P(D) none of these

Drawing the force diagram and the shear diagram,

Therefore, (A) is correct.

!

1

2P


13-4d2Mechanics of MaterialsBeamsExample 3 (FEIM):

For the shear diagram shown, what is the maximum bending moment? Thebending moment at the ends is zero, and there are no concentrated couples.(A) 8 kN • m(B) 16 kN • m(C) 18 kN • m(D) 26 kN • m

Starting from the left end of the beam, areas begin to cancel after 2 m. Startingfrom the right end of the beam, areas begin to cancel after 4 m. The rectangle onthe right has an area of 16 kN • m. The trapezoid on the left has an area of(1/2)(12 kN + 14 kN) (2 m) = 26 kN • m. The trapezoid has the largest bendingmoment.Therefore, (A) is correct.


13-4eMechanics of MaterialsBeams

Bending Stress

Deflection

Shear Stress

Note: Beam deflection formulas are given in the NCEES Handbook forany situation that might be on the exam.


13-4fMechanics of MaterialsBeams

Example (FEIM):

Find the tip deflection of the beam shown. EI is 3.47 × 106 N • m2, theload is 11 379 N/m, and the beam is 3.7 m long.

From the NCEES Handbook:

!

" =w

oL

4

8EI=

11379N

m

#

$ %

&

' ( 3.7 m( )

4

8( ) 3.47)106 N *m2( )

= 0.077 m


13-5aMechanics of MaterialsColumns

Beam-Columns (Axially Loaded Beams)

Maximum and minimum stresses in an eccentrically loaded column:


13-5bMechanics of MaterialsColumns

Euler’s Formula

Critical load that causes a long column to buckle:

r = the radius of gyrationk = the end-resistant coefficientkl = the effective length

= slenderness ratio

!

l

r


13-5cMechanics of MaterialsColumns

Elastic Strain Energy:

Strain energy per unit volume for tension:

Mechanics of Materials 13-1 - Valparaiso University

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