1. Stress is defined as force per unit area and can be calculated using the formula stress = force/area. The main types of stress are axial/normal stress, shear stress, and bearing stress. 2. Strain is the ratio of deformation to original length and is calculated using the formula strain = change in length/original length. Hooke's law states that stress is proportional to strain within the elastic limit defined by a material's Young's modulus. 3. Additional concepts covered include thin-walled pressure vessels, Poisson's ratio, thermal deformation, and stress-strain diagrams. Worked examples are provided to demonstrate calculating stresses, strains, deformations and other mechanical properties.

1. FUNDAMENTALS
OF
DEFORMABLE

2. STRESS
Stress is the expression of force applied to a
material per unit area of surface.
stress = force / area
𝜎 =
𝑃
𝐴

3. STRESS(S or σ) FORCE (P) AREA (A)
Psi or pounds/in2 (lb) Pound (lb) Sq.icnh (in2)
Pa or pascal (N/ m2) Newton(N) Sq. meter (m2)
MPa or megapascal
(N/mm2)
Newton(N) Sq. mm (mm2)

4. Axial Stress/ Normal Stress – is the stress
developed under the action of the forces
acting axially or perpendicular to the resisting
area.
Normal stress is either tensile stress or
compressive stress. Members subject to pure
tension (or tensile force) is under tensile stress,
while compression members (members subject to
compressive force) are under compressive
stress.

5. Compressive force will tend to shorten the
member. Tension force on the other hand will
tend to lengthen the member.

6. Ex. A steel bar is 10 mm in diameter and 2m
long. It is stretched with a force of 20 kN.
Calculate the stress carried.
Ans. 254.64 MPa
Ex. A hollow steel tube with an inside diameter of
100 mm must carry a tensile load of 400 kN.
Determine the outside diameter of the tube if the
stress is limited to 120 MN/m2 .
Ans. D = 119.35 mm

7. SHEARING STRESS
Forces parallel to the area resisting the force
cause shearing stress. Shearing stress is also
known as tangential stress.
where V is the resultant shearing force which
passes which passes through the centroid of the
area A being sheared
𝜏 =
𝑉
𝐴

8. Types of shearing stress:
1. Single Shear – produced when the force to
tend slide a single section
2. Double Shear – produced when force tend to
slide two sections.
3. Direct or Punching Shear – occurs over an
area parallel to applied load.
4. Induced Shear – occurs a section inclined w/
the resultant load.

10. Ex. What force is required to punch a 20-mm-
diameter hole in a plate that is 25 mm thick? The
shear strength is 350 MN/m2.
Ans. P = 549.8 kN
Ex. A single bolt is used to lap joint two steel bars
together. Tensile force on the bar is 8 kN. Solve the
diameter of the bolt required if the allowable
shearing stress on it is 70 MPa.
Ans. D = 11.65 mm

11. BEARING STRESS
Bearing stress is the contact pressure between
the separate bodies. It differs from compressive
stress, as it is an internal stress caused by
compressive forces.
𝜎𝑏 =
𝑃𝑏
𝐴𝑏

13. Ex. A 20 mm diameter rivet joints two plates which
are each 100 mm wide. If the allowable stresses
are 140 MPa for bearing in the plate material and
80 MPa for shearing of the rivet, determine the
minimum thickness of each plate.
Ans. t = 8.98 mm

14. Ex. Three 20 mm diameter rivets are used to lap
joint two steel plates together. Each plate is 25 mm
thick. The plates carry a tensile load of kN. Solve
the bearing stress in each plate. Assume the axial
load is distributed equally on the rivets.
Ans. 33.33 MPa

15. THIN-WALLED PRESSURE VESSELS
A tank or pipe carrying a fluid or gas under a
pressure is subjected to tensile forces, which resist
bursting, developed across longitudinal and
transverse sections.
Food for the brain: If the cylindrical tank is really thin-
walled, it is not important which diameter is used.
Although the inner diameter is used by common
convention.

16. Thin walled cylindrical tank
1. Tangential Stress/Circumferential Stress/Hoop
Stress/Girth Stress
-The stress that w/c acts tangent to the surface
of the cylinder.

17. If there exist an external pressure 𝑝𝑜
and an internal pressure 𝑝𝑖, the formula may
be expressed as:
𝑆𝑡 =
𝑝𝐷
2𝑡 𝑆𝑡 =
(𝑝𝑖 − 𝑝𝑜)𝐷
2𝑡

18. 2. Longitudinal Stress
- the bursting pressure is acting over the end of
the transversal section of the vessel the stress is
called longitudinal since it is acting parallel to the
longitudinal axis of vessel (cylinder)

19. If there exist an external pressure 𝑝𝑜
and an internal pressure 𝑝𝑖, the formula may
be expressed as:
𝑆𝐿 =
𝑝𝐷
4𝑡 𝑆𝐿 =
(𝑝𝑖 − 𝑝𝑜)𝐷
4𝑡

20. It can be observed that the tangential stress is
twice that of the longitudinal stress.
Note: For a cylindrical tank, the tangential stress will give
the bigger thickness, thus with pressure and stress given,
this will be the formula used in solving for the thickness
of a given cylindrical tank.
𝑆𝑡 = 2 𝑆𝐿

21. Relationship between pressure to height or depth:
𝑝 = 𝛾ℎ → 𝑆𝐺 =
𝛾
𝛾𝑤𝑎𝑡𝑒𝑟
𝑝 = 𝑆𝐺 𝛾𝑤𝑎𝑡𝑒𝑟 (ℎ)
Where:
𝑝 = pressure (at the bottom of the tank)
𝛾 = specific weight of liquid
ℎ = height of the free liquid surface from the bottom of the tank
𝑆𝐺 = specific gravity of the liquid

22. 3. Spherical Shell
If a spherical tank of diameter D and thickness
t contains gas under a pressure of p, the stress at
the wall can be expressed as:
𝑆𝑆 =
𝑝𝐷
4𝑡
𝑆𝑆 =
(𝑝𝑖 − 𝑝𝑜)𝐷
4𝑡

23. Ex. A cylindrical steel pressure vessel is fabricated from
steel plates which have thickness of 20 mm. The
diameter of the pressure vessel is 500 mm and its length
is 3 m. Determine the maximum internal pressure which
can be applied if the stress in the steel is limited to 140
MPa.
Ans. P = 11.2 MPa

24. Ex. A water tank is 8 m in diameter and 12 m high. If the
tank is completely filled, determine the minimum
thickness of the tank plating if the stress is limited to 40
Mpa.
Ans. t = 11.7 mm

25. Ex. Commercial propane stored in a spherical steel tank
generates a gage pressure of 160 psi. If the tank is 4 ft in
diameter and has a walls 0.25 inch thick, what maximum
tensile stress in psi is developed in the steel?
Ans. S = 7,680 psi

26. SIMPLE STRAIN
- Also known as unit deformation, strain is
the ratio of the change in length caused by the
applied force, to the original length. Units of
strain are in/in, mm/mm, percentage or unit
less.
𝜀 =
𝛿
𝐿
Where:
𝜀 = strain
𝛿 = elongation
L = length

27. Stress – Strain Diagram

28. 1. Proportional limit – is the maximum stress to w/c a
material can be subjected w/o any deviation from
the proportionality of stress and strain
2. Elastic Limit – is the maximum stress to w/c a material
may be subjected w/o trace of any permanent set
remaining upon complete withdrawal of stress.
3. Yield Point – is the stress w/c the stain begins to
increase very rapidly w/o a corresponding increase
in stress
4. Ultimate Strength - maximum stress to w/c a material
may be subjected before failure occurs.
5. Rapture Strength - is the strength of the material at
rupture. This is also known as the breaking
strength.

29. HOOKE’S LAW(ROBERT HOOKE)
Hooke's Law states that within the proportional
limit, the stress is directly proportional to strain.
But Thomas Young introduced a constant of
proportionality called Young’s Modulus or
commonly called Modulus of Elasticity.
𝜎 ∝ 𝜀 𝑜𝑟 𝜎 = 𝑘𝜀
𝜎 = 𝐸𝜀

30. Axial Deformation
- it is the deformation caused by axial loads
to the body
If not specified or given:
E for steel = 200 GPa or 30 x 106 psi
E for aluminum = 70 GPa
𝛿 =
𝑃𝐿
𝐴𝐸
=
𝜎𝐿
𝐸

31. Deformation due to its own weight
where ρ is in kg/𝑚3
, L is the length of the rod in mm,
m is the total mass of the rod in kg, A is the cross-
sectional area of the rod in 𝑚𝑚2
, and g = 9.81 m/𝑠2
.
𝛿 =
𝜌𝑔𝐿2
2𝐸
=
𝑚𝑔𝐿
2𝐴𝐸

32. Ex. A steel rod having a cross-sectional area of 300
mm2 and a length of 150 m is suspended vertically
from one end. It supports a tensile load of 20 kN at
the lower end. If the unit mass of steel is 7850
kg/m3 and E = 200 × 103 MN/m2, find the total
elongation of the rod.
Ans. 54.33 mm

33. Ex. A steel wire 30 ft long, hanging vertically,
supports a load of 500 lb. Neglecting the weight of
the wire, determine the required diameter if the
stress is not to exceed 20 ksi and the total
elongation is not to exceed 0.20 in. Assume E = 29
× 106 psi.
Ans. 0.1988 in

34. Statically Indeterminate Members
When the reactive forces or the internal
resisting forces over a cross section exceed the
number of independent equations of equilibrium,
the structure is called statically indeterminate.
These cases require the use of additional relations
that depend on the elastic deformations in the
members.

35. Ex. A steel bar 50 mm in diameter and 2 m long is surrounded
by a shell of a cast iron 5 mm thick. Compute the load that will
compress the combined bar a total of 0.8 mm in the length of 2
m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa.
Ans. P = 191.64 kN

36. Ex. A reinforced concrete column 200 mm in diameter is
designed to carry an axial compressive load of 300 kN.
Determine the required area of the reinforcing steel if the
allowable stresses are 6 MPa and 120 MPa for the
concrete and steel, respectively. Use Eco = 14 GPa and
Est = 200 GPa.
Ans. 1398.9 mm2

37. Poisson’s Ratio (𝝁)
– is the ratio of the magnitude of the lateral
strain to the magnitude of the longitudinal or axial
strain.
𝜇 =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛
=
𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙
𝜀𝑎𝑥𝑖𝑎𝑙

38. Food for the brain: Poisson’s Ratio is unit less material
property that never exceeds 0.5. If not given or specified, typical
values for steel, aluminum, and copper are 0.30, 0.33 and 0.34
respectively.
𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙 =
∆𝑑
𝑑
𝜀𝑎𝑥𝑖𝑎𝑙 =
∆𝐿
𝐿
=
𝛿
𝐿
where:
∆L = change in length
∆d = change in diameter

39. Food for the brain:
For a tensile force, volume
increase slightly while for
compressive force, volume
decrease slightly.
∆𝑉 = 𝑉(1 − 2𝜇)(𝜀𝑎𝑥𝑖𝑎𝑙)
where :
∆V = approximated value of the
change in volume
V = original volume
𝜇 = Poisson's ratio
𝜀𝑎𝑥𝑖𝑎𝑙 = axial strain

40. Ex. A square steel bar 50 mm on a side and 1 m
long is subjected to an axial tensile force of 250 kN.
Determine the decrease in lateral dimension due to
this load. Consider E = 200 GPa and 𝜇 = 0.3?
Ans. 0.0075 mm
Ex. What is the actual hole diameter by a steel
punch ¾ inch diameter subjected to a 40,000 lb
compressive load? Assume Poisson’s ratio of 0.3
Ans. 0.7507 in

41. Ex. A brass bar of length 2.25 m with a square
cross section of 90 mm on each side is subjected
to an axial tensile force of 1500 kN. Assume
Poisson’s ratio of 0.34 and E = 110 GPa.
Determine the approximate increase in the volume
of the bar.
Ans. 9818.18 mm3

42. Thermal Deformation
- if the temperature of an object is changed, the
object will experience length, area and volume
changes. The magnitude of these changes will
depend on the coefficient of linear thermal
expansion (α)

43. Note:
∆𝐿 = 𝛼𝐿 𝑇2 − 𝑇1 = ∆𝐿 = 𝛼𝐿 ∆𝑇
∆𝑉 = 𝛽𝑉 𝑇2 − 𝑇1 = 𝛽𝑉(∆𝑇)
where :
∆V = change in volume
V = original volume
∆L = change in length
L = original length
𝛼 = coefficient of linear expansion
𝛽 = coefficient of volume expansion
𝑇2 = final temperature
𝑇1 = initial temperature
𝛽 = 3𝛼
αsteel = 6.5 x10−6 / ℉
= 11.7 x10-6 / ℃

44. Ex. A brass sleeve of inside diameter of 1.9995 cm
and at 20ºC is to be heated so that it will just bearly
slide over a shaft of diameter of 2.0005 cm. To
what temperature must the sleeve be heated? For
brass, α = 1.9 x 10-6/ºC.
Ans. T = 283.223 ºC
Ex. The coefficient of linear expansion of glass is
9 x 10-6/ºC. If a specific bottle holds 50 mL at 15ºC,
find its capacity at 25ºC.
Ans. 50.0135 mL

45. If temperature deformation is permitted to
occur freely, no load or stress will be induced in the
structure. In some cases where temperature
deformation is not permitted, an internal stress is
created. The internal stress created is termed as
thermal stress.

46. For a homogeneous rod mounted between
unyielding supports as shown, the thermal stress is
computed as:
deformation due to temperature changes;
𝛿𝑇 = 𝛼𝐿∆𝑇

47. deformation due to equivalent axial stress;
where σ is the thermal stress in MPa and E is the
modulus of elasticity of the rod in MPa.
𝛿𝑃 =
𝑃𝐿
𝐴𝐸
=
𝜎𝐿
𝐸
𝛿𝑃 = 𝛿𝑇
𝜎𝐿
𝐸
= 𝛼𝐿∆𝑇
𝜎 = 𝐸𝛼∆𝑇

48. If the wall yields a distance of x as shown, the
following calculations will be made:
where σ represents the thermal stress.
𝛿𝑇 = 𝑥 + 𝛿𝑃
𝛼𝐿∆𝑇 = 𝑥 +
𝜎𝐿
𝐸

49. Ex. A steel rod with a cross-sectional area of 0.25
in2 is stretched between two fixed points. The
tensile load at 70°F is 1200 lb. What will be the
stress at 0°F? At what temperature will the stress
be zero? Assume α = 6.5 × 10-6 in / (in·°F) and E =
29 × 106 psi.
Ans.
𝜎 = 18 ksi
T = 95.46 °F

50. Ex. A steel rod is stretched between two rigid walls
and carries a tensile load of 5000 N at 20°C. If the
allowable stress is not to exceed 130 MPa at
-20°C, what is the minimum diameter of the rod?
Assume α = 11.7 μm/(m·°C) and E = 200 GPa.
Ans. d = 13.22 mm

51. Ex. Steel railroad reels 10 m long are laid with a
clearance of 3 mm at a temperature of 15°C. At
what temperature will the rails just touch? What
stress would be induced in the rails at that
temperature if there were no initial clearance?
Assume α = 11.7 μm/(m·°C) and E = 200 GPa.
Ans. 60 MPa

52. TORSION
The stress or Deformation caused when one end of an
object is twisted in one direction and the other end is held
motion less or twisted in opposite direction. Torsion on shaft
causes shearing strength.

53. For Solid Cylindrical For Hollow Cylindrical
Shaft: Shaft:
τ =
16T
πd3 τ =
16TD
π(D4 − d4)
Where:
τ = maximum shear stress
D = outside diameter
d = inside diameter
T = Torque

54. POWER TRANSMITTED BY THE SHAFT
P =
2πNT
60
P =
2πNT
33000
Where:
N = rotational speed (rpm)
T = Torque (N-m)
P = Power (watts)
Where:
N = rotational speed (rpm)
T = Torque (lb-ft)
P = Power (Hp)
P = Tω
Where:
ω = angular velocity (rad/s)
T = Torque (N-m)
P = Power (watts)

55. MAXIMUM TWISTED ANGLE IN THE SHAFT’S
FIBER
When the shaft is rotating, the shaft fiber will be
twisted at some angle.
Formula Mnemonic:
True Love / Just God
θ =
𝑇𝐿
𝐽𝐺 Where:
𝐿 = Length of the shaft (m)
T = Torque (N-m)
J = Polar Moment of Inertia
G = Modulus of Rigidity
(Shear Modulus)

56. Food for the brain:
IF NOT GIVEN : G for steel = 80 GPa or 11.5 x 106 psi
For Solid Shaft: For Hollow Shaft:
𝐽 =
𝜋
32
𝐷4
𝐽 =
𝜋
32
(𝐷4
−𝑑4
)

57. Ex. A steel shaft 3 ft long that has a diameter of 4 in. is subjected
to a torque of 15 kips - ft. Determine the maximum shearing
stress and the angle of twist. Use G = 12 × 106 psi.
Solution:
Given:
L = 3 ft D = 4 in
G = 12 × 106 psi T = 15 kips – ft
τ = ? θ = ?

58. τ =
16𝑇
𝜋𝐷3 =
16(15000 𝑙𝑏−𝑓𝑡)
12 𝑖𝑛
1 𝑓𝑡
𝜋(4 𝑖𝑛)3
τ = 14 323.94 psi
𝛕 = 14.32 ksi
𝜃 =
𝑇𝐿
𝐽𝐺
=
15000 𝑙𝑏−𝑓𝑡
12 𝑖𝑛
1 𝑓𝑡
(3 𝑓𝑡)
12 𝑖𝑛
1 𝑓𝑡
𝜋
32
4 𝑖𝑛 4(12 × 106 𝑝𝑠𝑖)
*J =
𝜋
32
𝐷4
𝜃 = 0.0215 rad
𝜽 = 1.23°

59. Ex. A steel marine propeller shaft 14 in. in diameter
and 18 ft long is used to transmit 5000 Hp at 189
rpm. If G = 12 × 106 psi, determine the maximum
shearing stress.
Ans. 3094.6 psi

60. Ex. A hollow steel shaft is designed to transmit 500 hp at
1800 rpm. If the outside diameter of the shaft is 3 inches
, how large may the inside diameter be so as not to
exceed a shearing stress of 10,000 psi?
Ans. T = 17,507.04 lb-in

61. Ex. A hollow steel shaft is designed to transmit 500 hp at
1800 rpm. If the outside diameter of the shaft is 3 inches
, how large may the inside diameter be so as not to
exceed a shearing stress of 10,000 psi?
Ans. d = 2.714 in

62. Ex. What is the minimum diameter of a solid steel shaft
that will not twist through more than 3 degrees in a 6 m
length when subjected to a torque of 14 kN-m? Use G =
83 Gpa
Ans. d = 118 mm

63. Ex. A steel shaft is 60 m long and has an outer diameter
of 340mm and inner diameter of 260 mm. If the shaft
delivers a power output 4.5 MW when the shaft rotates at
20 rad/s, determine its angle of twist. Assume G = 75
GPa for the shaft.
Ans. 11.95°

64. FLANGED BOLT COUPLINGS
- A flanged bolt couplings is a driving coupling
between rotating shaft that consists of flanges (or
half couplings) one of which is fixed at the end of
each shaft, the two flanges being bolted together
with a ring of bolts to complete the drive

65. 𝑇 = 𝑃𝑅𝑛
𝜏 =
𝑃
𝐴
; 𝑃 = 𝜏A
𝐴 =
𝜋𝐷2
4
Where:
𝑑 = diameter of each bolt A = area of each
n = number of bolts bolt
R = Radius of bolt circle
𝜏 = shearing stress on bolts
P = resisting force

66. FOR COUPLING WITH TWO CONCENTRIC
BOLT RADIUS
𝑇 = 𝑃1𝑅1𝑛1 + 𝑃2𝑅2𝑛2
Where:
where the subscript 1 refer to bolts on the
outer circle an subscript 2 refer to bolts on the inner
circle.

67. For rigid flanges, the shear deformations in the bolts are
proportional to their radial distances from the shaft axis.
The shearing strains are related by
Using Hooke’s law for shear, G =
𝜏
𝛾
, we have
𝛾1
𝑅1
=
𝛾2
𝑅2
𝜏1
𝐺1𝑅1
=
𝜏2
𝐺2𝑅2
𝑜𝑟
𝑃1
𝑅1
𝐺1𝑅1
=
𝑃2
𝑅2
𝐺2𝑅2

68. If the bolts on the two circles have the same area,
𝐴1 = 𝐴2, and if the bolts are made of the same
material, 𝐺1 = 𝐺2, the relation between 𝑃1 and 𝑃2
reduces to;
𝑃1
𝑅1
=
𝑃2
𝑅2
Where:
𝑃1 > 𝑃2

69. Ex. A flanged bolt coupling consists of ten 20-mm
diameter bolts spaced evenly around a bolt circle 400
mm in diameter. Determine the torque capacity of the
coupling if the allowable shearing stress in the bolts is 40
MPa.
Solution:
𝑇 = 𝑃𝑅𝑛 = 𝜏𝐴𝑅𝑛 = 𝜏
𝜋𝐷2
4
𝑅𝑛
𝑇 = (40𝑀𝑃𝑎)
𝜋 20𝑚𝑚 2
4
(200𝑚𝑚)(10)
𝑇 = 25.13 × 106
𝑁 − 𝑚𝑚
𝑇 = 25.13 kN − m

70. Ex. A flanged bolt coupling consists of ten steel ½ -in.-
diameter bolts spaced evenly around a bolt circle 14 in. in
diameter. Determine the torque capacity of the coupling if the
allowable shearing stress in the bolts is 6000 psi.
Ans. 6872.23 lb-ft
Ex. A torque of 700 lb-ft is to be carried by a flanged bolt
coupling that consists of eight ½ -in.-diameter steel bolts on a
circle of diameter 12 in. and six ½ -in.-diameter steel bolts on
a circle of diameter 9 in. Determine the shearing stress in the
bolts.
Ans. 626.87 psi outer circle
470.15 psi inner circle

71. HELICAL SPRING
- When close-coiled helical spring, composed
of a wire of round rod of diameter d wound into a
helix of mean radius R with n number of turns, is
subjected to an axial load P produces the following
stresses and elongation.

72. APPROXIMATION FORMULA:
𝜏 =
16𝑃𝑅
𝜋𝑑3
1 +
𝑑
4𝑅
WAHL’s FORMULA:
(more precise)
𝜏 =
16𝑃𝑅
𝜋𝑑3
4𝑚 − 1
4𝑚 − 4
+
0.615
𝑚
Where:
𝑚 =
2𝑅
𝑑

73. ELONGATION FORMULA:
δ =
64𝑃𝑅3
𝑛
𝐺𝑑4
Where:
𝑑 = diameter of spring wire
n = number of turns
R = mean Radius of the spring
𝜏 = shearing stress on spring
P = applied force or load carried
δ = elongation
G = modulus of rigidity
Spring Constant (k):
𝑘 =
𝑃
𝛿
=
𝐺𝑑4
64𝑅3𝑛

74. SPRINGS IN SERIES
For two or more springs with spring laid in series,
the resulting spring constant k is given by:
where 𝑘1, 𝑘2,… are the
spring constants for
different springs.

75. SPRINGS IN PARALLEL
where 𝑘1, 𝑘2,… are the
spring constants for
different springs.

76. Ex. Determine the maximum shearing stress and
elongation in a helical steel spring composed of 20 turns
of 20-mm-diameter wire on a mean radius of 90 mm
when the spring is supporting a load of 1.5 kN. Use
Wahl’s Eq. and G = 83 GPa.
Solution:
𝜏 =
16𝑃𝑅
𝜋𝑑3
4𝑚−1
4𝑚−4
+
0.615
𝑚
𝑚 =
2𝑅
𝑑
=
2(90 )
20
= 9
𝜏 =
16(1500 𝑁)(90 𝑚𝑚)
𝜋(20 𝑚𝑚)3
4(9)−1
4(9)−4
+
0.615
9
𝜏 = 99.87 MPa

77. δ =
64𝑃𝑅3𝑛
𝐺𝑑4 =
64(1500 𝑁) 90 𝑚𝑚 3(20)
(83 ×103 𝑀𝑃𝑎)(20 𝑚𝑚)4
δ = 105.4 𝑚𝑚

78. Ex. A helical spring is made by wrapping steel wire 20
mm diameter around a forming cylinder 150 mm in
diameter. Compute the number of turns required to
permit an elongation of 100 mm without exceeding a
shearing stress of 140 MPa. Use G = 83 GPa.
Ans. n = 17.89 turns

79. SHEAR AND MOMENT DIAGRAMS

80. Shear & Moment in Beams
DEFINITION OF A BEAM
A beam is a bar subject to forces or couples
that lie in a plane containing the longitudinal of the
bar. According to determinacy, a beam may be
determinate or indeterminate.

81. STATICALLY DETERMINATE BEAMS
Statically determinate beams are those beams in
which the reactions of the supports may be
determined by the use of the equations of static
equilibrium. The beams shown below are examples
of statically determinate beams.

83. STATICALLY INDETERMINATE BEAMS
If the number of reactions exerted upon a beam
exceeds the number of equations in static equilibrium, the
beam is said to be statically indeterminate. In order to solve
the reactions of the beam, the static equations must be
supplemented by equations based upon the elastic
deformations of the beam.
The degree of indeterminacy is taken as the
difference between the number of reactions to the number of
equations in static equilibrium that can be applied. In the
case of the propped beam shown, there are three reactions
𝑅1, 𝑅2, and M and only two equations (ΣM = 0 and sum;Fv =
0) can be applied, thus the beam is indeterminate to the first
degree (3 – 2 = 1).

85. TYPES OF LOADING
Loads applied to the beam may consist of a
concentrated load (load applied at a point), uniform load,
uniformly varying load, or an applied couple or moment.
These loads are shown in the following figures.

87. PROPERTIES OF SHEAR AND MOMENT DIAGRAMS
The following are some important properties of shear and moment
diagrams:
1. The area of the shear diagram to the left or to the right of the
section is equal to the moment at that section.
2. The slope of the moment diagram at a given point is the shear at
that point.
3. The slope of the shear diagram at a given point equals the load at
that point.
4. The maximum moment occurs at the point of zero shears. This is in
reference to property number 2, that when the shear (also the slope
of the moment diagram) is zero, the tangent drawn to the moment
diagram is horizontal.
5. When the shear diagram is increasing, the moment diagram is
concave upward.
6. When the shear diagram is decreasing, the moment diagram is
concave downward.

88. Ex. Draw the shear and moment diagrams for the
beams specified in the following problems. Give
numerical values at all change of loading positions
and at all points of zero shear.
Beam loaded as shown in Fig. P-425

89. 60 kN
30 kN
2 m 4 m 1 m
𝑅1 𝑅2
𝑀𝐴 = 0
60 2 + 30 7 = 𝑅2(6)
𝑅2 = 55 kN
𝐹𝑉 = 0
𝑅1+𝑅2= 60 + 30
𝑅1 = 60 + 30 − 55
𝑅1 = 35 kN
𝐴 𝐵 𝐶
𝐷

90. 60 kN
30 kN
2 m 4 m 1 m
𝑅1 = 35 𝑘𝑁 𝑅2 = 55 𝑘𝑁
To draw the Shear Diagram:
1) 𝑉𝐴 = 𝑅1 = 35 kN
2) 𝑉𝐵 = 𝑉𝐴 − 60 = 35 − 60
𝑉𝐵 = −25 kN
3) 𝑉𝐶 = 𝑉𝐵 + 55 = −25 + 55
𝑉𝐶 = 30 kN
4) 𝑉𝐷 = 𝑉𝐶 − 30 = 30 − 30
𝑉𝐷 = 0 kN
𝐴 𝐵 𝐶
𝐷
35 kN
-25 kN
30 kN
SHEAR DIAGRAM

91. 60 kN
30 kN
2 m 4 m 1 m
𝑅1 = 35 𝑘𝑁 𝑅2 = 55 𝑘𝑁
To draw the Moment
Diagram:
1) 𝑀𝐴 = 0
2) 𝑀𝐵 = 𝑀𝐴 + Area in shear Diagram
𝑀𝐵 = 0 + 35(2)
𝑀𝐵 = 70 kN − m
3) 𝑀𝐶 = 𝑀𝐵 + Area in shear Diagram
𝑀𝐶 = 70 − 4(25)
𝑀𝐶 = −30 kN − m
4) 𝑀𝐷 = 𝑀𝐶 − 30 = 30 − 30
𝑀𝐷 = 0 kN
𝐴 𝐵 𝐶
𝐷
35 kN
- 25 kN
30 kN
SHEAR DIAGRAM
MOMENT DIAGRAM
70 kN-m
- 30 kN-m
0 0

92. Ex. Draw the shear and moment diagrams for the beams
specified in the following problems. Give numerical values at all
change of loading positions and at all points of zero shear.
Cantilever beam acted upon by a uniformly distributed load and
a couple as shown in Fig. P-426.

93. 2 m 2 m 1 m
M = 60 kN-m
5 kN-m
To draw the Shear Diagram:
1) 𝑉𝐴 = 0
2) 𝑉𝐵 = 𝑉𝐴 + Area in Load Diagram
𝑉𝐵 = 0 − 5(2)
𝑉𝐵 = −10 kN − m
3) 𝑉𝐶 = 𝑉𝐵 + Area in Load Diagram
𝑉𝐶 = −10 + 0
𝑉𝐶 = −10 kN − m
4) 𝑉𝐷 = 𝑉𝐶 + Area in Load Diagram
𝑉𝐷 = −10 + 0
𝑉𝐷 = −10 kN − m
A B C D
- 10 kN-m
SHEAR DIAGRAM

94. 2 m 2 m 1 m
M = 60 kN-m
5 kN-m
A B C D
- 10 kN-m
To draw the Moment Diagram:
1) 𝑀𝐴 = 0
2) 𝑀𝐵 = 𝑀𝐴 + Area in shear Diagram
𝑀𝐵 = 0 −
1
2
(2)(10)
𝑀𝐵 = −10 kN − m
3) 𝑀𝐶 = 𝑀𝐵 + Area in shear Diagram
𝑀𝐶 = −10 − 2(10)
𝑀𝐶 = −30 kN − m
𝑀𝐶2 = −30 + 60 = 30 kN − m
4) 𝑀𝐷 = 𝑀𝐶2 +Area in shear Diagram
𝑀𝐷 = 30 − 1(10)
𝑀𝐷 = 20 kN − m
SHEAR DIAGRAM
MOMENT DIAGRAM
30 kN-m
20 kN-m
- 30 kN-m
- 10 kN-m
parabolic

95. Ex. Draw the shear and moment diagrams for the beams
specified in the following problems. Give numerical
values at all change of loading positions and at all points
of zero shear.
Beam loaded as shown in Fig. P-427.

96. Ex. Draw the shear and moment diagrams for the beams
specified in the following problems. Give numerical values at all
change of loading positions and at all points of zero shear.
Beam loaded as shown in Fig. P-428.

97. SHEAR STRESS OF THE BEAM
𝑆𝑠 =
3𝑉
2𝐴
→ for “rectangular” cross – section beam
𝑆𝑠 =
4𝑉
3𝐴
→ for “circular” cross – section beam
Where:
𝑉 = maximum shear force in the beam
A = cross sectional area of the beam
𝑆𝑠 = shear stress on beam

98. Ex. A round 1.5 cm diameter rod, 10 m long is
loaded a point load of 300 N at midspan. Calculate
the shear stress at neutral axis.
Solution:
Since symmetrically loaded,
𝑅1 = 𝑅2:
𝑅1 = 𝑅2=
300
2
= 150 𝑁
𝑅2
10 m
𝑅1
300 N
5 m

99. Draw the shear diagram:
1.) 𝑉𝐴 = 150 𝑁
2.) 𝑉𝐵 = 𝑉𝐴 − 300 = 150 − 300
𝑉𝐵 = −150 𝑁
3.) 𝑉
𝑐 = 𝑉𝐵 + 150 = −150 + 150
𝑉
𝑐 = 0
𝑅2 = 150 𝑁
10 m
𝑅1 = 150 𝑁
300 N
5 m
A B C
150 N
- 150 N
0
SHEAR DIAGRAM

100. Note: Maximum shear is 150 N
𝑆𝑆 =
4𝑉
3𝐴
→ for “circular” cross –
section beam
𝑆𝑆 =
4𝑉
3𝐴
→ 𝐴 =
𝜋
4
𝑑2
𝑆𝑆 =
4(150 𝑁)
3
𝜋
4
(15 𝑚𝑚)2
𝑆𝑆 = 1.13 𝑀𝑃𝑎
𝑅2 = 150 𝑁
10 m
𝑅1 = 150 𝑁
300 N
5 m
A B C
150 N
- 150 N
0
SHEAR DIAGRAM

101. BENDING STRESS OR FLEXURAL STRESS ON
BEAMS
𝑐
𝑁𝑒𝑢𝑡𝑟𝑎𝑙
𝑎𝑥𝑖𝑠
𝑆𝑏 =
𝑀𝑐
𝐼
→ 𝑍 =
𝐼
𝑐
Where:
𝑆𝑏 = bending stress
M = maximum moment on beam
I = moment of inertia of the cross -
sectional area of the beam
with respect to the center
c = distance of the neutral axis to
the extreme fiber
Z = section modulus

102. Bending Stress is the stress caused by the bending
moment of the beam.
𝐼 =
1
12
𝑏ℎ3 → for “rectangular” cross – section beam
𝐼 =
1
4
𝜋𝑟4
→ for “circular” cross – section beam

103. Ex. The cross section of the beam as shown below
is 5 inches wide and 10 inches deep. Determine
the maximum bending and shearing stresses?
Solution:
19 ft
𝑅1 𝑅2
2000 lbs
7 ft 7 ft
5 ft 300 lbs/ft
2000 lbs
300(14) = 4200 lbs
𝑅1 𝑅2
7 ft
12 ft
Simplify first the uniformly distributed
load to single point load.

104. 2000 lbs
300(14) = 4200 lbs
𝑅1 𝑅2
7 ft
12 ft
Solve for 𝑅1 and 𝑅2:
𝑀𝑅1
= 0
2000 + 4200 12 = 𝑅2(19)
𝑅2 = 3915.79 𝑙𝑏𝑠
𝐹𝑉 = 0
𝑅1 + 𝑅2 = 2000 + 4200
𝑅1 = 6200 − 3915.79
𝑅1 = 2284.21 𝑙𝑏𝑠

105. 19 ft
𝑅2 = 3915.79 𝑙𝑏𝑠
2000 lbs
7 ft 7 ft
5 ft 300 lbs/ft
Draw the shear diagram:
1.) 𝑉𝐴 = 2284.21 𝑙𝑏𝑠
2.) 𝑉𝐵 = 𝑉𝐴
𝑉𝐵 = 2284.21 𝑙𝑏𝑠
3.) 𝑉𝐶 = 𝑉𝐵 − 300(7)
𝑉𝐶 = 2284.21 − 300(7)
𝑉𝐶 = 184.21 𝑙𝑏𝑠
4.) 𝑉𝐶2 = 𝑉𝐶 − 2000
𝑉𝐶2 = 184.21 − 2000
𝑉𝐶2 = −1815.79 𝑙𝑏𝑠
5.) 𝑉𝐷 = 𝑉𝐶2 − 300 7
𝑉𝐷 = −3915.79 𝑙𝑏𝑠
6.) 𝑉𝐷2 = 𝑉𝐷 + 3915.79 = 0
A
B C D
2284.21 lbs
184.21 lbs
-1815.79 lbs
- 3915.79
lbs
0
SHEAR DIAGRAM
Maximum shear, V = 3915.79 lbs
𝑅1 = 2284.21 𝑙𝑏𝑠

106. 19 ft
2000 lbs
7 ft 7 ft
5 ft 300 lbs/ft
Draw the moment diagram:
1.) 𝑀𝐴 = 0
2.) 𝑀𝐵 = 𝑀𝐴 + Area in shear Diagram
𝑀𝐵 = 0 + 2284.21(5)
𝑀𝐵 = 11421.05 𝑙𝑏 − 𝑓𝑡
3.) 𝑀𝐶 = 𝑀𝐵 + Area in shear Diagram
𝑀𝐶 = 11421.05 +
1
2
(2284.21 + 184.21)(7)
𝑀𝐶 = 20060. 52 𝑙𝑏 − 𝑓𝑡
4.) 𝑀𝐷 = 𝑀𝐶 + Area in shear Diagram
𝑀𝐷 = 20060.52 −
1
2
(1815.79 + 3915.76)(7)
𝑀𝐷 = 0
Note: Moment is maximum when shear is
zero
A
B C D
0
SHEAR DIAGRAM
𝑅2 = 3915.79 𝑙𝑏𝑠
2284.21 lbs
184.21 lbs
-1815.79 lbs
- 3915.79
lbs
𝑅1 = 2284.21 𝑙𝑏𝑠
0 0
11421.05 lb - ft
20060.52 lb - ft
Maximum moment, M = 20060.52 lb - ft

107. For shearing stress on beam:
𝑆𝑠 =
3𝑉
2𝐴
→ for “rectangular” cross – section beam
𝑆𝑠 =
3𝑉
2𝐴
=
3(3915.79 𝑙𝑏𝑠)
2(5 𝑖𝑛)(10 𝑖𝑛)
𝑆𝑠 = 117.47 𝑝𝑠𝑖
For bending stress on beam:
𝑆𝑏 =
𝑀𝑐
𝐼
→ 𝐼 =
1
12
𝑏ℎ3
𝑆𝑏 = 2888.71 𝑝𝑠𝑖
𝑆𝑏 =
𝑀𝑐
1
12
𝑏ℎ3
=
12(20060.52 𝑙𝑏 −𝑓𝑡)
12 𝑖𝑛
1 𝑓𝑡
(5 𝑖𝑛)
(5 𝑖𝑛)(10 𝑖𝑛)3

108. Ex. A 6 cm simply supported beam carries a uniformly
distributed load of 1 kN/m and a concentrated load of 2 kN
at the middle. Find the maximum bending stress. The beam
is 5 cm wide and 10 cm deep.
Ans. 90 MPa
Ex. The allowable bending stress of the beam shown below
is 24000 psi. What is the required section modulus?
Ans. 7.5 in3
Ex. The beam shown is 10 inches wide and 20 inches high.
Determine (a) bending stress and (b) shear stress midway
between supports.
Ans. (a) 574 psi
(b) 2.01 psi

109. CANTILEVER BEAMS
- is a long projecting beams fixed only at one
end.

110.  Point load P at end of a cantilever beam
𝛿
𝐿
𝑃
𝑀
𝑀 = 𝑃𝐿 𝛿 =
𝑃𝐿3
3𝐸𝐼
Where:
M = maximum moment I = centroidal moment of inertia
𝛿 = maximum beam deflection
L = length
E = modulus of elasticity

111.  Point load P at any location on a cantilever
beam
𝛿
𝐿
𝑃
𝑀
𝑀 = 𝑃𝑏 𝛿 =
𝑃𝑏2
6𝐸𝐼
(3𝐿 − 𝑏)
𝑎
𝑏

112.  Couple C at end of a cantilever beam
𝛿
𝐿
𝐶
𝑀
𝑀 = 𝐶 𝛿 =
𝐶𝐿2
2𝐸𝐼

113.  Uniform distributed load along the entire
length
𝛿
𝐿
𝑀
𝑀 =
𝑤𝐿2
2
𝛿 =
𝑤𝐿4
8𝐸𝐼
𝑤

114.  Uniform distributed load along the entire
length
𝛿
𝐿
``
𝑀
𝑀 =
𝑤𝐿2
6
𝛿 =
𝑤𝐿4
30𝐸𝐼
𝑤

115. Ex. A steel cantilever beam 16 ft and 8 inches in length is
subjected to a concentrated load of 320 lbs acting at the
free end of the bar. The beam is of rectangular cross
section, 2 inches wide by 3 inches deep. Determine the
magnitude of the maximum bending stress of the beam.
Solution:
𝐿 = 16 ′ 12′′
𝑃 = 320 𝑙𝑏𝑠
𝑀

116. 𝐿 = 16 𝐹𝑡
12 𝑖𝑛
1 𝐹𝑡
+ 8 𝑖𝑛
𝐿 = 200 𝑖𝑛𝑐ℎ𝑒𝑠
𝑀 = 𝑃𝐿 = (320 𝑙𝑏)(200 𝑖𝑛)
𝑀 = 64 000 𝑙𝑏 − 𝑖𝑛
𝑆𝑏 =
𝑀𝑐
𝐼
where: 𝐼 =
1
12
𝑏ℎ3
𝑆𝑏 =
12𝑀𝑐
𝑏ℎ3 =
12(64000 𝑙𝑏−𝑖𝑛)(1.5 𝑖𝑛)
(2 𝑖𝑛)(3 𝑖𝑛)3
𝑆𝑏 = 21 333.33 psi
𝐿 = 16 ′ 8′′
𝑃 = 320 𝑙𝑏𝑠
𝑀
𝑐. 𝑔.
𝑐 = 1.5′′
𝑏 = 2′′
ℎ = 3′′

117. Ex. A cantilever beam 3 m long is subjected to a
uniform distributed load of 30 kN/m length. The
cross section is rectangular, 110 mm wide and 220
mm deep. Determine the maximum bending stress.
Solution:
𝑀
𝑤 = 30 𝑘𝑁/𝑚
𝐿 = 3𝑚
𝑐 = 110 𝑚𝑚
𝑏 = 110 𝑚𝑚
ℎ =
220 𝑚𝑚
𝑐. 𝑔.

118. 𝑀 =
𝑤𝐿2
2
=
(30 𝑘𝑁/𝑚)(3𝑚)2
2
𝑀 = 135 𝑘𝑁 − 𝑚 = 135 𝑀𝑁 − 𝑚𝑚
𝑆𝑏 =
𝑀𝑐
𝐼
where: 𝐼 =
1
12
𝑏ℎ3
𝑆𝑏 =
12𝑀𝑐
𝑏ℎ3
=
12(135 𝑀𝑁−𝑚𝑚)(110 𝑚𝑚)
(110 𝑚𝑚)(220 𝑚𝑚)3
𝑆𝑏 = 152.14 MPa

119. Ex. What would be the maximum stress in a
cantilever beam holding a mass of 5000 kg at its
end if it is 3 m long and is made of a 200 mm wide
by 300 high wooden beam?
Ans. 49 Mpa
Ex. A cantilever 8 foot, 3.5’’ x 3.5’’ timber is loaded
with a 200 lb point load at the end. What is the
maximum deflection at the end? E = 1700 ksi
Ans. 2.78 in