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7/21/20

Chapter 1: Stress

Introduction• Mechanics of materials is a study of the relationship

between the external loads on a body and the

ntens ty o t e n erna oa s w t n t e o y.

• This subject also involves the deformations and

stability of a body when subjected to external forces.

© 2010 Mechanical & Manufacturing, © 2010 Mechanical & Manufacturing, UniKLUniKL--MFIMFI



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Equilibrium of a Deformable Body

External ForcesExternal Forces

1.Surface Forces

- caused by direct contact

of other body’s surface

2.Body Forces

- other body exerts a force


without contact


ReactionsReactions

Surface forces developed at the supports/points of

contact between bodies.




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Equations of EquilibriumEquations of Equilibrium

Equilibrium of a body requires a balance of forces and a balance of moments

For a body with x , y , z coordinate system with origin

O,

0M 0F == ∑∑ O

0 ,0 ,0 === ∑∑∑ z y x F F F


Best way to account for these forces is to draw the body’s free-body diagram (FBD).

,, === z y x


Internal Resultant LoadingsInternal Resultant Loadings

Objective of FBD is to determine the resultant force

and moment acting within a body.

In general, there are 4 different types of resultant

loadings:

a) Normal force, N

b) Shear force, V


c) Torsional moment or torque, T

d) Bending moment, M



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Example 1.1Determine the resultant internal loadings acting on the cross section at C of thebeam.

Solution:Free body Diagram

mN1809

270

6=⇒= w

w

Distributed loading at C is found by proportion,


Magnitude of the resultant of the distributed load,

( )( )N5406180

2

1 ==F

which acts from C ( ) m263

1 =

Solution:Equations of Equilibrium

(Ans)0

0 ;0

=

=−=→+ ∑C

C x

N

N F

Applying the equations of equilibrium we have


( )

(Ans)mN0108

02540 ;0

(Ans)540

0540 ;0

⋅−=

=−−=+

=

=−=↑+

∑

∑

C

C C

C

C y

M

M M

V

V F



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Example 1.5Determine the resultant internal loadings acting on the cross section at B of thepipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of

50 N and a couple moment of 70 N·m at its end A. It is fixed to the wall at C .


SolutionFree-Body Diagram

( )( )( )

( )( )( ) N525.2481.925.12

N81.981.95.02

==

==

AD

BD

W

W

Calculating the weight of each segment of pipe,

Applying the six scalar equations of equilibrium,

( )

( )

( )

( ) (Ans) N3.84

050525.2481.9 ;0

(Ans) 0 ;0

(Ans) 0 ;0

=

=−−−=

==

==

∑∑∑

x B

z B z

y B y

x B x

F

F F

F F

F F

( ) ( ) ( ) ( ) ( ) 025.081.95.0525.245.05070 ;0 =−−−+=∑ x B x B M M


( )( ) ( ) ( ) ( )

( )

( ) ( ) (Ans) 0 ;0

(Ans) mN8.77

025.150625.0525.24 ;0

(Ans) mN3.30

==

⋅−=

=++=

⋅−=

∑

∑

z B z B

y B

y B y B

x B

M M

M

M M

M



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Stress

Distribution of internal loading is important in

mechanics of materials.

We will consider the material to be continuous .

This intensity of internal force at a point is called

stress .


StressNormal StressNormal Stress σσ

Force per unit area acting normal to ∆A

Shear StressShear Stress ττ

Force per unit area acting tangent to ∆A

A

F z A

z∆

∆=

→∆ 0limσ


A

F

A

y

A zy

x A

zx

∆

∆=

∆=

→∆

→∆

0

0

lim

lim

τ

τ



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Average Normal Stress in an Axially Loaded Bar

When a cross-sectional area bar is subjected to

axial force through the centroid, it is only subjected

to norma stress.

Stress is assumed to be averaged over the area.


Average Normal Stress in an Axially Loaded Bar

Average Normal Stress DistributionAverage Normal Stress Distribution

When a bar is subjected to a

constant deformation,

EquilibriumEquilibrium A

P

AP

dAdF A

=

=

= ∫ ∫

σ

σ

σ

σ = average normal stress

P = resultant normal force

A = cross sectional area of bar


2 normal stress components

that are equal in magnitude

but opposite in direction.



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Example 1.6The bar has a constant width of 35 mm and a thickness of 10 mm. Determine themaximum average normal stress in the bar when it is subjected to the loading

shown.

Solution:By inspection, different sections have different internal forces.


Graphically, the normal force diagram is as shown.

Solution:

By inspection, the largest loading is in region BC ,

kN30= BC P

-


,

the largest average normal stress is

( )( )( )

(Ans)MPa7.8501.0035.0

10303

=== A

P BC BC σ



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3kN/m80=st γ

Example 1.8The casting is made of steel that has a specific weight of

. Determine the average compressive stress

acting at points A and B .

Solution:By drawing a free-body diagram of the top segment,

the internal axial force P at the section is

( )( ) ( ) 02.08.080

0 ;0

2

=

=−

=−=↑+ ∑P

W PF st z

π


.

The average compressive stress becomes

( )(Ans) kN/m0.64

2.0

042.8 2

2===

π σ

A

P

Average Shear Stress The average shear stress distributed over each

sectioned area that develops a shear force.

2 different types of shear:

A

V avg

=τ

τ = average shear stress

P = internal resultant shear force

A = area at that section


a) Single Shear b) Double Shear



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1

Example 1.12The inclined member is subjected to a compressive force of 3000 N. Determinethe average compressive stress along the smooth areas of contact defined by AB

and BC , and the average shear stress along the horizontal plane defined byEDB .

Solution:The compressive forces acting on the areas of contact are


( )

( ) N240003000 ;0

N180003000 ;0

5

4

5

3

=⇒=−=↑+

=⇒=−=→+

∑∑

BC BC y

AB AB x

F F F

F F F

The shear force acting on the sectioned horizontal plane EDB is

Solution:

N1800 ;0 ==→+ ∑ V F x

Average compressive stresses along the AB and BC planes are

( )( )

( )( )(Ans) N/mm20.1

4050

2400

(Ans) N/mm80.14025

1800

2

2

==

==

BC

AB

σ

σ


( )( )(Ans) N/mm60.0

4075

1800 2==avg

τ

Average shear stress acting on the BD plane is



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Allowable Stress

Many unknown factors that influence the actual

stress in a member.

A factor of safety is needed to obtained allowable

load.

The factor of safety (F.S.) is a ratio of the failure

load divided by the allowable load

allow

fail

F

F SF =.


allow

fail

allow

fail

SF

SF

τ

τ

σ

σ

=

=

.

.

Example 1.14The control arm is subjected to the loading. Determine to the nearest 5 mm the

required diameter of the steel pin at C if the allowable shear stress for the steel is

. Note in the figure that the pin is subjected to double shear.MPa55=allowableτ

Solution:For equilibrium we have


( ) ( ) ( )( )

( )

( ) kN3002515 ;0

kN502515 ;0

kN150125.025075.0152.0 ;0

5

3

5

4

5

3

=⇒=−−=+↑

=⇒=+−−=+→

=⇒=−==+

∑∑∑

y y y

x x x

AB ABC

C C F

C C F

F F M



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1

Solution:The pin at C resists the resultant force at C . Therefore,

) ) kN41.3030522 =−=F

m1045.2761055

205.15 26

3

2

×=×

== −V A

allowableτ

The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-

sectional area between the arm and each supporting leaf for the pin.

The required area is


mm8.18

mm45.2462

2

=

=⎟ ⎠

⎜⎝

d

d π

Use a pin with a diameter of d = 20 mm. (Ans)

Example 1.17The rigid bar AB supported by a steel rod AC having a diameter of 20 mm and an

aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter

pins at A and C are subjected to single shear . If the failure stress for the steel and

aluminum is and res ectivel and the failure( ) MPa680=ist σ ( MPa70=σ

shear stress for each pin is , determine the largest load P that can be

applied to the bar. Apply a factor of safety of F.S. = 2.

Solution:The allowable stresses are

( )MPa340

680=== failst σ

σ

ail

MPa900= fail


( )( )

MPa4502

900

..

MPa352

70

..

2..

===

===

SF

SF

SF

fail

allow

failal

allowal

a ows

τ τ

σ σ



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There are three unknowns and we apply the equations of equilibrium,

Solution:

( ) ( ) (1) 0225.1 ;0 =−=+∑ F P M AC B

. =−=B A

We will now determine each value of P that creates the allowable stress in the rod,

block, and pins, respectively.

For rod AC, ( ) ( ) ( ) ( ) kN8.10601.01034026 === π σ AC allowst AC AF

Using Eq. 1,( )( )

kN17125.1

28.106==P


For block B, ( ) ( ) ( )[ ] kN0.63101800103566 === −

Ballowal B AF σ

Using Eq. 2, ( )( ) kN16875.0

20.63==P

Solution:

For pin A or C, ( ) ( ) kN5.114009.01045026 ==== π τ AF V allow AC

Using Eq. 1,( )( )

kN18325.114

==P25.1

When P reaches its smallest value (168 kN), it develops the allowable normal

stress in the aluminium block. Hence,

(Ans)kN168=P


ch01 [compatibility mode]

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