MECHANICS OF MATERIALS 7 th Edition

Russell C. Hibbeler

Chapter 1: Stress

© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd

Chapter 1: StressChapter 1: StressMechanics of Material 7Mechanics of Material 7thth Edition Edition

Introduction• Mechanics of materials is a study of the relationship

between the external loads on a body and the intensity of the internal loads within the body.

• This subject also involves the deformations and stability of a body when subjected to external forces.



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




ReactionsReactionsSurface forces developed at the supports/points of

contact between bodies.




Equations of EquilibriumEquations of EquilibriumEquilibrium 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,

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

0M 0F O

0 , 0 , 0

0 , 0 , 0

zyx

zyx

MMM

FFF




Internal Resultant LoadingsInternal Resultant LoadingsObjective 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



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

Solution:Free body Diagram

mN1809

270

6 w

wDistributed loading at C is found by proportion,

Magnitude of the resultant of the distributed load,

N540618021 F

which acts from C m2631



Solution:Equations of Equilibrium

(Ans) mN 0108

02540 ;0

(Ans) 540

0540 ;0

(Ans) 0

0 ;0

C

CC

C

Cy

C

Cx

M

MM

V

VF

N

NF

Applying the equations of equilibrium we have



Example 1.5Determine the resultant internal loadings acting on the cross section at B of the pipe. 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

N 525.2481.925.12

N 81.981.95.02

AD

BD

W

W

Calculating the weight of each segment of pipe,

Applying the six scalar equations of equilibrium,

(Ans) N 3.84

050525.2481.9 ;0

(Ans) 0 ;0

(Ans) 0 ;0

xB

zBz

yBy

xBx

F

FF

FF

FF

(Ans) 0 ;0

(Ans) mN8.77

025.150625.0525.24 ;0

(Ans) mN3.30

025.081.95.0525.245.05070 ;0

zBzB

yB

yByB

xB

xBxB

MM

M

MM

M

MM



StressDistribution 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 Stress Normal Stress σσForce per unit area acting normal to ΔA

Shear StressShear Stress ττForce per unit area acting tangent to ΔA

A

FzA

z

0

lim

A

FA

F

y

Azy

x

Azx

0

0

lim

lim



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 normal stress.

Stress is assumed to be averaged over the area.



Average Normal Stress in an Axially Loaded Bar

Average Normal Stress DistributionAverage Normal Stress DistributionWhen a bar is subjected to a

constant deformation,

EquilibriumEquilibrium2 normal stress components

that are equal in magnitude but opposite in direction.

A

P

AP

dAdFA

σ = average normal stressP = resultant normal forceA = cross sectional area of bar



Example 1.6The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum 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,

kN 30BCP

Since the cross-sectional area of the bar is constant, the largest average normal stress is

(Ans) MPa 7.85

01.0035.0

1030 3

A

PBCBC



3kN/m 80st

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

kN 042.8

02.08.080

0 ;02

P

P

WPF stz

The average compressive stress becomes

(Ans) kN/m 0.64

2.0

042.8 22

A

P



Average Shear StressThe average shear stress distributed over each

sectioned area that develops a shear force.

2 different types of shear:

A

Vavg

τ = average shear stressP = internal resultant shear forceA = area at that section

a) Single Shear b) Double Shear



Example 1.12The inclined member is subjected to a compressive force of 3000 N. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB.

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

N 240003000 ;0

N 180003000 ;0

54

53

BCBCy

ABABx

FFF

FFF



The shear force acting on the sectioned horizontal plane EDB is

Solution:

N 1800 ;0 VFx

Average compressive stresses along the AB and BC planes are

(Ans) N/mm 20.14050

2400

(Ans) N/mm 80.14025

1800

2

2

BC

AB

(Ans) N/mm 60.04075

1800 2avg

Average shear stress acting on the BD plane is



Allowable StressMany 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

allow

fail

allow

fail

SF

SF

F

FSF

.

.

.



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.

Solution:For equilibrium we have

MPa 55allowable

kN 3002515 ;0

kN 502515 ;0

kN 150125.025075.0152.0 ;0

53

54

53

yyy

xxx

ABABC

CCF

CCF

FFM



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

kN 41.30305 22 CF

mm 8.18

mm 45.2462

m 1045.2761055

205.15

2

263

2

d

d

VA

allowable

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

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 respectively, and the failure 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

MPa 680failst

MPa 4502

900

..

MPa 352

70

..

MPa 3402

680

..

SF

SF

SF

failallow

failal

allowal

failst

allowst

MPa 70failalMPa 900fail



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

Solution:

(2) 075.02 ;0

(1) 0225.1 ;0

PFM

FPM

BA

ACB

We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.

For rod AC, kN 8.10601.010340 26 ACallowstAC AF

Using Eq. 1, kN 171

25.1

28.106P

For block B, kN 0.631018001035 66 BallowalB AF

Using Eq. 2, kN 168

75.0

20.63P



Solution:

For pin A or C, kN 5.114009.010450 26 AFV allowAC

Using Eq. 1,

kN 18325.1

25.114P

When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,

(Ans) kN 168P

MECHANICS OF MATERIALS 7 th Edition

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