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EAT203

Lecture 1: Stress-strain relationship

EAT203, Engineering Mechanics

Lecture Topic Tutorial

1 Stress-strain relationship

2 Shear force (SF) and bending moment (BM) for non-uniform loading 1

3 SF and BM for statically indeterminate problem

4 Shear stress in bending 2

5 Stress transformation

6 Experimental stress / strain 3

7 Yield criteria (theories of failure)

8 Stress concentration + Fatigue 4

9 Vibration (Forced with damping) 5

10 Kinematics - particle (velocity diagram)

11 Kinematics - ridig body (velocity diagram)

12 Kinematics - acceleration diagram 6

13 Gear system

14 Balancing of rotating system 7

[1] Benham, P.P., Crawford, R.J. & Armstrong, C.G., Mechnics of Engineering Materials,

2nd ed., Prentice Hall, UK (1996)

[2] Hibbeler, R. C., Mechanics of Materials, 7th ed., Prentice Hall, Singapore (2008)

[3] Hibbeler, R. C., Engineering Mechanics Dynamics, 8th ed., Prentice-Hall International

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Stress

• The effect of external applied forces on a

solid body or the member of a framework can

be measured in terms of the internal reacting

forces.

• The intensity of internal force at a point is

called stress.

• Stress is defined as the internal force per

unit cross section area at right angle to the

direction of the force.

Average Normal Stress Distribution

When a bar is subjected to a constant deformation

A

P

AP

dAdFA

σ = average normal stress

P = resultant normal force

A = cross sectional area of bar

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Shear Stress,

• If an applied load consists of two equal and

opposite forces which are not in the same line,

then the material being loaded will tend to

shear as shown.

• Shear stress is the force (acting tangent to the

area) per unit area

A

F

Strain

Direct strain

• Consider a uniform bar subjected to an axial

tensile load, F. If the resulting extension of

the bar is and its unloaded (original) length

is L, then the direct tensile strain is

L

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Shear strain

• A shear stress produces a shear strain as is

shown below. The shear strain is defined as

• For small deflection,

• The shear strain is dimensionless and is

measured in radians

yx

tan

Example 1

The plate is deformed into the dashed shape. If, in this

deformed shape, horizontal lines on the plate remain

horizontal and do not change their length, determine (a)

the average normal strain along the side AB, and (b) the

average shear strain in the plate relative to the x and y

axes.

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Line AB, coincident with the y axis, becomes line AB’

after deformation, thus the length of this line is

The average normal strain for AB is

The negative sign indicates the strain causes a

contraction of AB.

Solution: Part (a)

mm 018.24832250' 22AB

(Ans) 1093.7

250

250018.248'

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AB

ABABavgAB

As noted, the once 90°angle BAC between the sides of

the plate, referenced from the x, y axes, changes to θ’

due to the displacement of B to B’.

Since then is the angle shown in the

figure.

Solution: Part (b)

'2

xy xy

(Ans) rad 0121.0

2250

3tan 1

xy

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G

Hooke’s Law defines the linear relationship

between the normal stress and strain within

the elastic region, i.e., Young’s modulus.

Eσ = normal stress

E = Young’s modulus

ε = normal strain

While Young’s modulus describe the material’s

response to direct strain, shear modulus, G

describe the material’s response to shear strain

vGEand 12

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• states that in the elastic range, the lateral

strain is proportional to the direct strain (due

to stress), where

• Negative sign, as direct strain

(negative) causes lateral

expansion (positive strain),

and vice versa.

• Typical values for Poisson’s ratio are 1/3 or

1/4.

direct

lateralv

Poisson’s Ratio, v

Example 2

A steel bar (E = 210GPa, v = 0.32) has the dimensions

shown. If an axial force of is applied to the bar, determine

the change in its length and the change in the dimensions

of its cross section after applying the load. The material

behaves elastically.

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Solution:

The normal stress in the bar is

6

9

6

107610210

100.16 st

zz

E

Pa 100.1605.01.0

1080 63

A

Pz

The axial elongation of the bar is therefore

(Ans) m1145.11076 6

z

zz L

The contraction strains in x and y directions are

-66 103.24107632.0

zstyx v

The changes in the dimensions of the cross section are

(Ans) m215.105.0103.24

(Ans) m43.21.0103.24

6

6

yyy

xxx

L

L

9

zyx

zyxx

E

v

E

E

v

E

v

E

zx

y

yE

v

E

3D stress-strain

xy

zz

E

v

E

yzyxyv

E

vv

vE

1211

xzyxxv

E

vv

vE

1211

zzyxzv

E

vv

vE

1211

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Plane stress

E

v

E

yxx

E

v

E

xy

y

yxz

E

v

Coefficient of Linear Thermal Expansion

• Most materials increase in volume as their

temperature increases

• Change in length of a structural member of

length L is proportional to temperature

increase T:

= a.T.L

• Constant of proportionality a (K-1) is

coefficient of linear thermal expansion

{C.T.E.}

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• Increasing temperature causes expansion

and thus a positive strain and vice versa.

• Then thermal strain, = /L = a.T

L

A B

Unconstrained Bar

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Axial Stress in Constrained Bar

• If there is any restriction on the material, then

a thermal stress will result. For an initially

unstressed prismatic bar AB, an axial stress

(F) developed as it undergoes an increase in

temperature.

L

L

F

A

A

B

B

• In this case, principle of superposition is to

be applied

• Within elastic limit, assess effects of

temperature change and load separately

• Add results, noting sign, to determine total

effect, i.e. total strain = sum of strains due to

both external loads and thermal strains

Total strain, = strain due to external load, F + thermal strain

0 = -(F/EA) + a.T

F = a.T.E.A

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Example 3