ENG204 – Strength of MaterialsLecture 8, Sections 4.6, 4.12, 4.14Bending of Composites- Consider a member composed of two

materials.- Since the stress-strain relationship was

not considered in the derivation of the axial deformation, εx still varies linearly with y distance with bending.

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ρε y

x −=

- However, this time we can no longer assume that the N.A. passes through the centroid of the section.

- Our goal now is the find the N.A.- So, first find σx in each material ( )

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ρεσ yEE xx −==

- The forces exerted on elements of area dA in the upper and lower portions are

- If we let n = E2 / E1 , then

- Note that dF2 is the same force that would be exerted on an area ndA in the first material. i.e. resistance of bar will remain the same if lower portion is widened (or narrowed) by a factor of n. New section is called transformed section.

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dAyEdAdFρ

σ 111 −==

dAyEdAdFρ

σ 222 −==

)()( 112 ndAyEdAynEdF

ρρ−==

- Now the N.A. can be located by finding the centroid of the transformed section.

- If follows that normal stress can be found using the elastic flexure formula using I of the transformed section.

- Important: σx computed is given in terms of E1. To get σx for E2, it must be multiplied by n!

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IMy

x =σ

- Deformation of a composite can also be determined using the transformed section by writing the curvature as

- Where I is the moment of inertia of the transformed section.

- An important example of structural members made of two different materials is reinforced concrete.

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IEM

1

1=

ρ

Eccentric Loading- Thus far, we dealt with axial forces that pass through

the N.A. (centric loading) or moment forces (bending loading) separately.

- These two types of forces can act jointly to create eccentric loadings.

- The principle of superposition can be used to analyze effects of eccentric loadings.

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bendingxcentricxx )()( σσσ +=

- We note that stress for centric loading is

- and stress for bending moment is

- So the combined effect is , or

- Note that the combined effect can have same or different signs for the overall stress.

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bendingxcentricxx )()( σσσ +=

AP

centricx =)(σ

IMy

bendingx −=)(σ

IMy

AP

x −=σ

General Case of Eccentric Loading- Eccentric loading could cause bending in two directions

when a problem is analyzed in 3D—about y and z axes (two axes of the plane of the cross-section). In that case, the general form of stress equation must be used.

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y

y

z

zx I

zMI

yMAP

+−=σ

- To find N.A., set σx to zero, so that

- Reorder to get

- which is the equation of a line representing the N.A.

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y

y

z

z

IzM

IyM

AP

+−=0

z

z

zy

yz

AMPIz

MIMI

y +=

Problem 4.39- A steel bar (Es = 210 GPa) and aluminum bar (Ea = 70

GPa) are bonded together to form the composite bar shown. Determine the maximum stress in (a) The aluminum (b) the steel, (c) the radius of curvatures, when the bar is bend about the horizontal axis with M = 60 N. m.

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Problem 4.40- A steel bar (Es = 210 GPa)

and aluminum bar (Ea = 70 GPa) are bonded together to form the composite bar shown. Determine the maximum stress in (a) The aluminum (b) the steel, (c) the radius of curvatures, when the bar is bend about the horizontal axis with M = 60 N. m.

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Problem 4.57- A steel pipe (Es = 210

GPa) and an aluminum pipe (Ea = 70 GPa) are securely bonded together to form the composite beam shown. If beam is bent by M = 500 N.m, determine the maximum stress in (a) the aluminum (b) the steel.

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Example 1- Knowing that

the magnitude of the horizontal force Pis 8 kN, determine the stress at (a) point A, (b) point B.

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Problem 4.122- An eccentric force P is

applied as shown to a steel bar of 25 x 90-mm cross section. The strains at A and B have been measured and found to be εA = +350 µ, εB = -70 µ. Knowing that E = 200 GPa, determine (a) the distance d, (b) the magnituede of the force P.

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Example 2- A horizontal load P is

applied as shown to a short section of an S250 x 37.8 rolled-steel member. Knowing that the compressive stress in the member is not to exceed 82 MPa, determine the largest permissible load P.

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120mm

38mm

S250 x 37.8

Example 3- A vertical 4.80 kN load

is applied as shown on the wooden post of rectangular cross section, 80 x 120 mm. (a) determine the stress at points A, B, C, and D. (b) locate the neutral axis of the cross section.

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