1. Internal Forces
Stress
Lesson #1

2. Analysis of
Internal
Forces

3. Internal Forces
forces that act on cross sections
that are internal to the body
itself

4. 4
Analysis of Internal Forces
Resultant Force
Resultant
Force-Couple

5. 5
Analysis of Internal Forces
Shear Force
Axial/Normal Force

6. 6
Analysis of Internal Forces
Normal / Axial Force
Tension Force
Compressive Force

7. 7
Analysis of Internal Forces
Torque
Bending Moment

8. 8
Analysis of Internal Forces
Where;
P: The component of the resultant force that is perpendicular to the cross section,
tending to elongate or shorten the bar, is called the normal force.
V: The component of the resultant force lying in the plane of the cross section,
tending to shear (slide) one segment of the bar relative to the other segment, is
called the shear force.
T: The component of the resultant couple that tends to twist (rotate) the bar is
called the twisting moment or torque.
M: The component of the resultant couple that tends to bend the bar is called the
bending moment.

9. 9
Deformations produced by the components
of internal forces and couples.

10. 10
Deformations produced by the components
of internal forces and couples.

11. 11
Deformations produced by the components
of internal forces and couples.

12. 12
Deformations produced by the components
of internal forces and couples.

13. Stress
13

14. Stress
14
Normal
stress
Shear
stress
Bearing
stress

15. Stress
15
Shear component of R
Normal Component of R
Part of the resultant force
that is transmitted across A

16. Stress
16
Shear component of stress
Normal Component of stress
Stress Vector 𝑡 = lim
∆𝐴→0
∆𝑅
∆𝐴
𝜎 = lim
∆𝐴→0
∆𝑃
∆𝐴
=
𝑑𝑃
𝑑𝐴
𝜏 = lim
∆𝐴→0
∆𝑉
∆𝐴
=
𝑑𝑉
𝑑𝐴

17. 17
If the stresses are uniformly distributed,
•𝜎 =
𝑃
𝐴
Normal/Axial
Stress
• =
𝑉
𝐴
Shear Stress

18. 18
If the stresses are uniformly distributed,
Normal / Axial
Stress
CompressionStress
Tension Stress

19. 19
Units
Stress(,)
Pascal (Pa)
Mega
Pascal(Mpa)
Kip/in2 (ksi)
Lb/in2 (psi)
Force(P,V)
Newton (N)
kips
pounds
Area
M2
mm2
ft2
in2

20. 20
Axially Loaded Bars
Centroidal (axial) Loading
Normal Stress
𝜎 =
𝑃 → 𝑁𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝐴 → 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎

21. 21
⊷ Problem#3
The bar ABC consists of 2 cylindrical segments with different lengths
and cross-sectional areas. Axial loads are applied as shown. Calculate
the normal stress in each segment.
4000 𝑙𝑏 9000 𝑙𝑏
1.3 𝑓𝑡
1.6 𝑓𝑡
1.2𝑖𝑛2
1.8𝑖𝑛2
𝑨 𝑩
𝑪

22. 22
⊷ Problem#3
The bar ABC consists of 2 cylindrical segments with different
lengths and cross-sectional areas. Axial loads are applied as
shown. Calculate the normal stress in each segment.
4000 𝑙𝑏 𝑃𝐴𝐵
𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐴𝐵

23. 23
⊷ Problem#3
The bar ABC consists of 2 cylindrical segments with different
lengths and cross-sectional areas. Axial loads are applied as
shown. Calculate the normal stress in each segment.
𝑃𝐵𝐶
𝑆𝑒𝑔𝑚𝑒𝑛𝑡 𝐵𝐶
4000 𝑙𝑏 9000 𝑙𝑏
1.3 𝑓𝑡
1.6 𝑓𝑡
1.2𝑖𝑛2
1.8𝑖𝑛2
𝑨 𝑩

24. 24
⊷ Problem#4
The figure below shows a two-member truss supporting a
block of weight W. The cross-sectional areas of the members
are 800 mm2 for AB and 400 mm2 for AC. Determine the
maximum safe value of W if the working stresses are 110 MPa
for AB and 120 MPa for AC.

25. 25
⊷ Problem#4
The figure below shows a two-member truss supporting a
block of weight W. The cross-sectional areas of the
members are 800 mm2 for AB and 400 mm2 for AC.
Determine the maximum safe value of W if the working
stresses are 110 MPa for AB and 120 MPa for AC.
𝑻𝑨𝑩
𝑻𝑨𝑪
𝑻𝑨𝑩
𝑻𝑨𝑪
𝒘
𝑇𝐴𝐵𝑠𝑖𝑛400
𝑇𝐴𝐵𝑐𝑜𝑠400
𝑇𝐴𝐶𝑠𝑖𝑛600
𝑇𝐴𝐶𝑐𝑜𝑠600

26. 26
⊷ Problem#4
The figure below shows a two-member truss supporting a
block of weight W. The cross-sectional areas of the
members are 800 mm2 for AB and 400 mm2 for AC.
Determine the maximum safe value of W if the working
stresses are 110 MPa for AB and 120 MPa for AC.
𝑻𝑨𝑩
𝑻𝑨𝑪
𝒘
𝑇𝐴𝐵𝑠𝑖𝑛400
𝑇𝐴𝐵𝑐𝑜𝑠400
𝑇𝐴𝐶𝑠𝑖𝑛600
𝑇𝐴𝐶𝑐𝑜𝑠600

27. 27
⊷ Problem#4
The figure below shows a two-member truss supporting a
block of weight W. The cross-sectional areas of the
members are 800 mm2 for AB and 400 mm2 for AC.
Determine the maximum safe value of W if the working
stresses are 110 MPa for AB and 120 MPa for AC.
𝑻𝑨𝑩
𝑻𝑨𝑪
𝒘
𝑇𝐴𝐵𝑠𝑖𝑛400
𝑇𝐴𝐵𝑐𝑜𝑠400
𝑇𝐴𝐶𝑠𝑖𝑛600
𝑇𝐴𝐶𝑐𝑜𝑠600

28. 28
Determine the smallest allowable cross-sectional areas of members CE, BE,
and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller to reduce the danger of
buckling.

29. 29
Determine the smallest allowable cross-sectional areas of members BC, BE,
and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller to reduce the danger of
buckling.
𝐹𝐵𝐶
𝐹𝐵𝐸 𝐹𝐵𝐸
𝐹𝐹𝐸
𝐹𝐹𝐸

30. 30
Determine the smallest allowable cross-sectional areas of members BC, BE,
and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller to reduce the danger of
buckling.
𝐹𝐵𝐶
𝐹𝐵𝐸 𝐹𝐵𝐸
𝐹𝐹𝐸
𝐹𝐹𝐸

31. 31
Determine the smallest allowable cross-sectional areas of members BC, BE,
and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller to reduce the danger of
buckling.
𝐹𝐵𝐶
𝐹𝐵𝐸 𝐹𝐵𝐸
𝐹𝐹𝐸
𝐹𝐹𝐸

32. 32
Determine the smallest allowable cross-sectional areas of members BC, BE,
and EF for the truss shown. The working stresses are 20 ksi in tension and 14 ksi in
compression. (The working stress in compression is smaller to reduce the danger of
buckling.
𝐹𝐵𝐶
𝐹𝐵𝐸 𝐹𝐵𝐸
𝐹𝐹𝐸
𝐹𝐹𝐸

33. Shear Stress
33

34. • is tangent to the plane on which it
acts
• it arises whenever the applied loads
cause one section of a body to slide
past its adjacent section
Shear Stress ()
𝜏 =
𝑉 → 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
𝐴 → 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎

35. 35
(a) Single shear in a
Rivet
(b) Double shear in a
Bolt
(c) shear in a metal
sheet produced by a
punch.

36. Bearing Stress
36

37. (a) a rivet in a lap joint (b) bearing stress is
not constant
(c) bearing stress
caused by the
bearing force Pb is
assumed to be
uniform on
projected
area td.

38. 38
If two bodies are
pressed against each
other, compressive
forces are developed
on the area of contact.
The pressure caused
by these surface loads
is called bearing stress.
Bearing Stress
𝜎𝑏 =
𝑃𝑏
𝐴𝑏
=
𝑃
𝑡𝑑

39. 39
PROBLEM #5: When the force P reached 8 kN,
the wooden specimen shown failed in shear along
the surface indicated by the dashed line. Determine
the average shearing stress along that surface at the
time of failure.

40. 40
PROBLEM #5: When the force P reached 8 kN, the wooden specimen shown
failed in shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.

41. 41
PROBLEM #6: The wooden members
A and B are to be joined by plywood
splice plates that will be fully glued on
the surfaces in contact. As part of the
design of the joint, and knowing that
the clearance between the ends of the
members is to be 1/4 inch, determine
the smallest allowable length 𝐿 if the
average shearing stress in the glue is
not to exceed 120 psi.

42. 42
PROBLEM #6: The wooden members A and B are to be joined by plywood splice plates that
will be fully glued on the surfaces in contact. As part of the design of the joint, and knowing that
the clearance between the ends of the members is to be 1/4 inch, determine the smallest
allowable length 𝐿 if the average shearing stress in the glue is not to exceed 120 psi.

43. 43
⊷ Problem#7
The lap joint is connected by three 20-mm-diameter rivets. Assuming
that the axial load P = 50 kN is distributed equally among the three
rivets, find (a) the shear stress in a rivet; (b) the bearing stress between
a plate and a rivet; and (c) the maximum average tensile stress in each
plate.

44. 44
Problem #7. The lap joint is connected by three 20-mm-diameter rivets. Assuming that the axial
load P = 50 kN is distributed equally among the three rivets, find (a) the shear stress in a rivet; (b) the
bearing stress between a plate and a rivet; and (c) the maximum average tensile stress in each plate.

45. 45
Problem #7. The lap joint is connected by three 20-mm-diameter rivets. Assuming that the axial
load P = 50 kN is distributed equally among the three rivets, find (a) the shear stress in a rivet; (b) the
bearing stress between a plate and a rivet; and (c) the maximum average tensile stress in each plate.

46. 46
Problem #7. The lap joint is connected by three 20-mm-diameter rivets. Assuming that the axial
load P = 50 kN is distributed equally among the three rivets, find (a) the shear stress in a rivet; (b) the
bearing stress between a plate and a rivet; and (c) the maximum average tensile stress in each plate.

47. 47
Problem #7. The lap joint is connected by three 20-mm-diameter rivets. Assuming that the axial
load P = 50 kN is distributed equally among the three rivets, find (a) the shear stress in a rivet; (b) the
bearing stress between a plate and a rivet; and (c) the maximum average tensile stress in each plate.

48. 48
PROBLEM #8: In the clevis
shown in the figure, find the
minimum bolt diameter and the
minimum thickness of each yoke
that will support a load P = 14
kips without exceeding a
shearing stress of 12 ksi and a
bearing stress of 20 ksi.

49. 49
PROBLEM #8: In the clevis shown in the figure, find the minimum bolt
diameter and the minimum thickness of each yoke that will support a load P
= 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress
of 20 ksi.

50. Stresses on
Inclined
Plane
50

51. 51
⊷ Stresses on Inclined Plane

52. 52
Normal Stress
Shear Stress

53. 53
Problem #9. The two pieces of wood, 2 in. by 4 in., are glued
together along the 400 joint. Determine the maximum safe
axial load P that can be applied if the shear stress in the glue is
limited to 250 psi.

54. 54
Problem #9. The two pieces of wood, 2 in. by 4 in., are glued together along
the 400 joint. Determine the maximum safe axial load P that can be applied if
the shear stress in the glue is limited to 250 psi.
𝑷
𝑷𝑵 𝑷𝑽
4"
2
sin 400

55. Working
Stress and
Factor of
Safety
55

56. Working stress 𝜎𝑤, also called the
allowable stress,
• is the maximum safe axial stress used in
design.
• the working stress should be limited to
values not exceeding the proportional limit
so that the stresses remain in the elastic
range.
𝜎𝑤 =
𝜎𝑦𝑝
𝑁
𝑜𝑟 𝜎𝑤 =
𝜎𝑢𝑙𝑡
𝑁
Where:
𝜎𝑦𝑝 = 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠
𝜎𝑢𝑙𝑡 = 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑁 = 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦

57. Two forces are applied to the bracket BCD as
shown. (a) Knowing that the control rod AB is
to be made of a steel having an ultimate
normal stress of 600 MPa, determine the
diameter of the rod for which the factor of
safety with respect to failure will be 3.3. (b)
The pin at C is to be made of a steel having
an ultimate shearing stress of 350 MPa.
Determine the diameter of the pin C for which
the factor of safety with respect to shear will
also be 3.3. (c) Determine the required
thickness of the bracket supports at C
knowing that the allowable bearing stress of
the steel used is 300 MPa.
Problem #2

58. 58

59. 59

60. 60

61. 61
⊷ Problem Set #1: Normal Stress
Solve Problem #1.4,1.6, 1.7, 1.8, 1.11, 1.12, 1.13,
1.16, 1.18, 1.19
Page: 25-28
Reference: Strength of Materials
Andrew Pytel. Jaan Kiusalaas

62. 62
⊷ Problem Set #2
Solve Problem #1.28,1.29, 1.31, 1.34, 1.37, 1.39,
1.43, 1.45, 1.54, 1.59,
Page: 31-36
Reference: Strength of Materials
Andrew Pytel. Jaan Kiusalaas