The document summarizes key concepts from Chapter 1 of Mechanics of Materials by Russell C. Hibbeler, including: 1) Mechanics of materials relates external loads on a body to internal stresses and deformations. External forces include surface and body forces. 2) Equilibrium requires balancing total forces and moments. Free body diagrams are used to determine resultant internal forces like normal stress, shear stress, torque, and bending moment. 3) Stress is defined as internal force intensity and can be normal stress or shear stress. Average stresses are determined based on loading and geometry. 4) A factor of safety is used to determine allowable stresses below theoretical failure values due to uncertainties.

1. Russell C. Hibbeler
Chapter 1: Stress

2. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th 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.

3. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Equilibrium of a Deformable Body
External Forces
1.Surface Forces
- caused by direct contact
of other body’s surface
2.Body Forces
- other body exerts a force
without contact

4. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Equilibrium of a Deformable Body
Reactions
 Surface forces developed at the supports/points of
contact between bodies.

5. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Equilibrium of a Deformable Body
Equations 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,
 Best way to account for these forces is to draw
the body’s free-body diagram (FBD).
0
M
0
F 
 
 O
0
,
0
,
0
0
,
0
,
0












z
y
x
z
y
x
M
M
M
F
F
F

6. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Equilibrium of a Deformable Body
Internal 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

7. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.1
Determine the resultant internal loadings acting on the cross section at C of the
beam.
Solution:
Free body Diagram
m
N
180
9
270
6


 w
w
Distributed loading at C is found by proportion,
Magnitude of the resultant of the distributed load,
   N
540
6
180
2
1


F
which acts from C
  m
2
6
3
1


8. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Solution:
Equations of Equilibrium
 
(Ans)
m
N
0
108
0
2
540
;
0
(Ans)
540
0
540
;
0
(Ans)
0
0
;
0























C
C
C
C
C
y
C
C
x
M
M
M
V
V
F
N
N
F
Applying the equations of equilibrium we have

9. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.5
Determine 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.

10. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Solution
Free-Body Diagram
   
    N
525
.
24
81
.
9
25
.
1
2
N
81
.
9
81
.
9
5
.
0
2




AD
BD
W
W
Calculating the weight of each segment of pipe,
Applying the six scalar equations of equilibrium,
 
 
 
  (Ans)
N
3
.
84
0
50
525
.
24
81
.
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
         
 
       
 
    (Ans)
0
;
0
(Ans)
m
N
8
.
77
0
25
.
1
50
625
.
0
525
.
24
;
0
(Ans)
m
N
3
.
30
0
25
.
0
81
.
9
5
.
0
525
.
24
5
.
0
50
70
;
0





















z
B
z
B
y
B
y
B
y
B
x
B
x
B
x
B
M
M
M
M
M
M
M
M

11. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
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.

12. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Stress
Normal Stress σ
 Force per unit area acting normal to ΔA
Shear Stress τ
 Force per unit area acting tangent to ΔA
A
Fz
A
z




 0
lim

A
F
A
F
y
A
zy
x
A
zx










0
0
lim
lim



13. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
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.

14. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Average Normal Stress in an Axially Loaded Bar
Average Normal Stress Distribution
 When a bar is subjected to a
constant deformation,
Equilibrium
 2 normal stress components
that are equal in magnitude
but opposite in direction.
A
P
A
P
dA
dF
A


 




σ = average normal stress
P = resultant normal force
A = cross sectional area of bar

15. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.6
The 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.

16. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Graphically, the normal force diagram is as shown.
Solution:
By inspection, the largest loading is in region BC,
kN
30

BC
P
Since the cross-sectional area of the bar is constant,
the largest average normal stress is
 
  
(Ans)
MPa
7
.
85
01
.
0
035
.
0
10
30 3



A
PBC
BC


17. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
3
kN/m
80

st

Example 1.8
The 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
0
2
.
0
8
.
0
80
0
;
0
2







 
P
P
W
P
F st
z

The average compressive stress becomes
 
(Ans)
kN/m
0
.
64
2
.
0
042
.
8 2
2





A
P

18. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
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 = average shear stress
P = internal resultant shear force
A = area at that section
a) Single Shear b) Double Shear

19. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.12
The 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
2400
0
3000
;
0
N
1800
0
3000
;
0
5
4
5
3
















BC
BC
y
AB
AB
x
F
F
F
F
F
F

20. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
The shear force acting on the sectioned horizontal plane EDB is
Solution:
N
1800
;
0 


  V
Fx
Average compressive stresses along the AB and BC planes are
  
  
(Ans)
N/mm
20
.
1
40
50
2400
(Ans)
N/mm
80
.
1
40
25
1800
2
2




BC
AB


  
(Ans)
N/mm
60
.
0
40
75
1800 2


avg

Average shear stress acting on the BD plane is

21. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
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
allow
fail
allow
fail
S
F
S
F
F
F
S
F







.
.
.

22. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.14
The 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
55

allowable

      
 
  kN
30
0
25
15
;
0
kN
5
0
25
15
;
0
kN
15
0
125
.
0
25
075
.
0
15
2
.
0
;
0
5
3
5
4
5
3



























y
y
y
x
x
x
AB
AB
C
C
C
F
C
C
F
F
F
M

23. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Solution:
The pin at C resists the resultant force at C. Therefore,
    kN
41
.
30
30
5
2
2



C
F
mm
8
.
18
mm
45
.
246
2
m
10
45
.
276
10
55
205
.
15
2
2
6
3
2












 
d
d
V
A
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)

24. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Example 1.17
The 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
680

fail
st

 
 
 
 
MPa
450
2
900
.
.
MPa
35
2
70
.
.
MPa
340
2
680
.
.









S
F
S
F
S
F
fail
allow
fail
al
allow
al
fail
st
allow
st






  MPa
70

fail
al

MPa
900

fail


25. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
There are three unknowns and we apply the equations of equilibrium,
Solution:
   
    (2)
0
75
.
0
2
;
0
(1)
0
2
25
.
1
;
0










P
F
M
F
P
M
B
A
AC
B
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
.
106
01
.
0
10
340
2
6


 
 AC
allow
st
AC A
F
Using Eq. 1,
   kN
171
25
.
1
2
8
.
106


P
For block B,      
  kN
0
.
63
10
1800
10
35 6
6


 
B
allow
al
B A
F 
Using Eq. 2,
   kN
168
75
.
0
2
0
.
63


P

26. © 2008 Pearson Education South Asia Pte Ltd
Chapter 1: Stress
Mechanics of Material 7th Edition
Solution:
For pin A or C,    
  kN
5
.
114
009
.
0
10
450
2
6



 
 A
F
V allow
AC
Using Eq. 1,
   kN
183
25
.
1
2
5
.
114


P
When P reaches its smallest value (168 kN), it develops the allowable normal
stress in the aluminium block. Hence,
(Ans)
kN
168

P