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Chapter 2.0
SIMPLE STRESSES
CE-MDB222 (Mechanics of Deformable Bodies)
Prepared by:
Engr. Bryan P. Abgao
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SIMPLE STRESSES
Stress, σ – is the internal resistance of a material to the distorting
effects of external forces.
Force per unit area; Unit strength
Units: MPa (SI), psi (English)
Types: Normal, Shearing, and Bearing
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SIMPLE STRESSES
Given two bars of same lengths but of different
materials, suspended from a common support,
determine which bar is stronger?
Material
Max. Load
Cross-sectional
Area
A
500 N
10 mm2
B
5000 N
1000 mm2
Ans.
Bar A is stronger due to higher
maximum allowable stress.
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2.1 NORMAL/AXIAL STRESS, σ
• stress caused by axial force (tensile or compressive)
• acts normal/perpendicular to the resisting surface.
• results in volume change
𝝈=
𝑷
𝑨
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Sample Problem 2.1
A hollow steel tube with an outside diameter of 200
mm is subjected to an axial tensile force of 400 kN as
shown in the figure.
Determine the safest thickness of the tube (mm) if
the stress is limited to 130 MPa.
Round off to the nearest whole number.
Ans.
𝒕 = 𝟔 𝒎𝒎
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Sample Problem 2.2
The bar ABCD in the figure shown consists of three cylindrical steel segments with
different lengths and cross-sectional areas. Axial loads are applied as shown.
Calculate the normal stress in each segment.
Ans.
𝝈𝑨𝑩 = 𝟑𝟑𝟑𝟑. 𝟑𝟑 𝒑𝒔𝒊 𝑻
𝝈𝑩𝑪 = 𝟐𝟕𝟕𝟕. 𝟕𝟖 𝒑𝒔𝒊 𝑪
𝝈𝑪𝑫 = 𝟒𝟑𝟕𝟓 𝒑𝒔𝒊 𝑪
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Sample Problem 2.3
Axial loads are applied to the compound rod that is composed of an aluminum
segment rigidly connected between steel and bronze segments. Find the largest
safe value of P (kN) if the working stresses are 120 MPa for steel, 68 MPa for
aluminum, and 110 MPa for bronze.
Ans.
𝑷 = 𝟏𝟐 𝒌𝑵
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Sample Problem 2.4
The column consists of a wooden post and a
concrete footing, separated by a steel bearing
plate. Find the maximum safe value of the axial
load P (lb) if the working stresses are 1000 psi for
wood and 450 psi for concrete.
Ans.
𝑷 = 𝟓𝟎, 𝟐𝟔𝟓. 𝟒𝟖 𝒍𝒃
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Sample Problem 2.5
Determine the mass of the heaviest uniform
cylinder that can be supported in the position
shown without exceeding a stress of 50 MPa in
cable BC. Neglect friction and the weight of bar
AB. The cross-sectional area of BC is 100 mm2.
Ans.
𝟔𝟏𝟏. 𝟔𝟐 𝒌𝒈
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Sample Problem 2.6
The uniform 300-lb bar AB carries a 500-lb vertical
force at A. The bar is supported by a pin at B and the
0.5-in diameter cable CD. Find the stress in the cable.
Ans.
𝟖, 𝟐𝟕𝟔. 𝟎𝟔 𝒑𝒔𝒊
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Sample Problem 2.7
The figure shows the landing gear of a light airplane.
Determine the compressive stress (MPa) in strut AB
caused by the landing reaction R = 40 kN. Neglect the
weights of the members. The strut is a hollow tube,
with 50-mm outer diameter and 40-mm inner
diameter.
Ans.
𝟗𝟒. 𝟑𝟏 𝑴𝑷𝒂
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Sample Problem 2.8
For the truss shown in the figure, calculate the normal stresses in member AC, and member
BD. The cross-sectional area of each member is 900 mm2.
Ans.
𝝈𝑨𝑪 = 𝟓𝟗. 𝟐𝟔 𝑴𝑷𝒂 𝑻
𝝈𝑩𝑫 = 𝟕𝟒. 𝟎𝟕 𝑴𝑷𝒂 𝑪
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Sample Problem 2.9
The figure shows a two-member truss
supporting a block of weight W. The crosssectional areas of the members are 800 mm2
for AB and 400 mm2 for AC.
Determine the maximum safe value of W
(kN) if the working stresses are 110 MPa for
AB and 120 MPa for AC.
Ans.
𝟔𝟏. 𝟕𝟏 𝒌𝑵
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Sample Problem 2.10
The 1000-kg uniform bar AB is suspended from two
cables AC and BD; each with cross-sectional area 400
mm2. Find the magnitude P (kN) and location x (m)
of the largest additional vertical force that can be
applied to the bar. The stresses in AC and BD are
limited to 100 MPa and 50 MPa, respectively.
Ans.
𝑷 = 𝟓𝟎. 𝟏𝟗 𝒌𝑵
𝒙 = 𝟎. 𝟔𝟎 𝒎
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2.2 SHEARING STRESS, 𝛕
• stress caused by shearing force (e.g. bolts, pins, rivets)
• acts tangent or parallel to the resisting surface
• results in shape change
𝝉=
𝑷𝒔
𝑽
=
𝑨𝒔 𝑨𝒔
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2.2 SHEARING STRESS, 𝛕
Examples of direct shear:
(a) Single shear in rivet
(b) Double shear in a bolt
(c) Shear in a metal sheet
produced by a punch
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2.2 SHEARING STRESS, 𝛕
Single shear failure
in pin specimens
double shear failure
In a pin specimen
punching shear failure cracks on
concrete slab (bottom face)
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2.2 SHEARING STRESS, 𝛕
Case 1: Single Shear
𝝉=
𝑽 𝑷
=
𝑨 𝑨
𝝉=
𝑽 𝑷/𝟐
𝑷
=
=
𝑨
𝑨
𝟐𝑨
Case 2: Double Shear
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Sample Problem 2.11
Determine the shear stress (MPa) in the 20-mm-diameter
pin at A and the 30-mm-diameter pin at B that support the
beam in the figure.
Ans.
𝝉𝑨 = 𝟑𝟒 𝑴𝑷𝒂
𝝉𝑩 = 𝟏𝟕. 𝟔𝟖 𝑴𝑷𝒂
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Sample Problem 2.12
The bell crank, which is in equilibrium under the forces shown in the figure, is supported by
a 20-mm-diameter pin at D that is in double shear. Determine (a) the required diameter
(mm) of the connecting rod AB, given that its tensile working stress is 100 MPa; (b) the
shear stress (MPa) in the pin.
Ans.
𝒅𝑨𝑩 = 𝟏𝟗. 𝟗𝟐 𝒎𝒎
𝝉𝑫 = 𝟖𝟒. 𝟑𝟑 𝑴𝑷𝒂
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Sample Problem 2.13
A circular plate as shown is to be punched in a plate
that has a shear strength of 40 ksi. The working
compressive stress for the punch is 50 ksi.
(a) Compute the maximum thickness of the plate in
which a 2.5-in.-diameter hole can be punched.
(b) If the plate is 0.25 in. thick, determine the
diameter of the smallest hole that can be
punched.
Ans.
(a) 𝟎. 𝟕𝟖 𝒊𝒏
(b) 𝟎. 𝟖𝟎 𝒊𝒏
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Sample Problem 2.14
The rectangular wood panel is formed by gluing
together two boards along the 30-degree seam as
shown in the figure. Determine the largest axial
force P (lb) that can be carried safely by the panel if
the working stress for the wood is 1120 psi, and the
normal and shear stresses in the glue are limited to
700 psi and 450 psi, respectively.
Ans.
𝑷 = 𝟑, 𝟕𝟑𝟑. 𝟑𝟑 𝒍𝒃
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Sample Problem 2.15
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 the shear stress (MPa) in a
rivet.
Ans.
𝝉 = 𝟓𝟑. 𝟎𝟓 𝑴𝑷𝒂
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Sample Problem 2.16
The plate welded to the end of the I-beam is fastened
to the support with four 10-mm-diameter bolts (two
each side). Assuming that the load is equally divided
among the bolts, determine the normal and shear
stresses (MPa) in a bolt.
Ans.
𝝈 = 𝟕𝟑. 𝟏𝟓 𝑴𝑷𝒂
𝝉 = 𝟔𝟏. 𝟑𝟖 𝑴𝑷𝒂
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2.3 BEARING STRESS, σb
• contact pressure between separate bodies.
• differs from normal stress which is the internal stress caused by a
compressive force.
• Examples are soil pressure beneath a support and the contact pressure
between a rivet and the side of its hole.
𝝈𝒃 =
𝑷𝒃
𝑨𝒃
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2.3 BEARING STRESS, σb
Examples of bearing stress:
(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 Ab = td.
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2.3 BEARING STRESS, σb
sample of bearing stress failures
at a bolted connection
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Sample Problem 2.17
The cylindrical steel column has an outer diameter
of 4 in. and inner diameter of 3.5 in. The column
is separated from the concrete foundation by a
square bearing plate. The working compressive
stress is 26000 psi for the column, and the
working bearing stress is 1200 psi for concrete.
Find the largest force P (lb) that can be applied to
the column.
Ans.
𝑷 = 𝟓𝟖, 𝟖𝟎𝟎 𝒍𝒃
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Sample Problem 2.18
The lap joint is fastened with four ¾-in-diameter
rivets. The working stresses are 14 ksi for the
rivets in shear and 18 ksi for the plates in bearing.
Find the maximum safe axial load P (kips) that can
be applied to the joint. Assume that the load is
equally distributed among the rivets.
Ans.
𝑷 = 𝟐𝟒. 𝟕𝟒 𝒌𝒊𝒑𝒔
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Sample Problem 2.19
Assume that the axial load P is applied to the lap joint is distributed equally among the
three 20-mm diameter rivets. What is the maximum load P (kN) that can be applied if
the allowable stresses are 40 MPa for shear in rivets, 90 MPa for bearing between a
plate and a rivet, and 120 MPa for tension in the plates?
Ans.
𝑷 = 𝟑𝟕. 𝟕𝟎 𝒌𝑵
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Sample Problem 2.20
The 20-mm-diameter bolt fastens two wooden
planks together. The nut is tightened until the
tensile stress in the bolt is 150 MPa. Find the
smallest safe diameter d of the washers if the
working bearing stress for wood is 13 MPa.
Ans.
𝒅 = 𝟕𝟎. 𝟖𝟐 𝒎𝒎
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Sample Problem 2.21
Three wood boards, each 4 in. wide, are joined by the
¾-in.-diameter bolt. If the working stresses for wood
are 800 psi in tension and 1500 in bearing, find the
largest allowable value of the force P (lb).
Ans.
𝑷 = 𝟐, 𝟐𝟓𝟎 𝒍𝒃
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Sample Problem 2.22
For the joint shown in the figure, calculate
(a) the largest bearing stress (MPa) between
the pin and the members; (b) the average
shear stress (MPa) in the pin; and (c) the
largest average normal stress (MPa) in the
members.
Ans.
𝝈𝒃 (𝒎𝒂𝒙) = 𝟏𝟔𝟔. 𝟔𝟕 𝑴𝑷𝒂
𝝉𝒑𝒊𝒏 = 𝟏𝟎𝟏. 𝟖𝟔 𝑴𝑷𝒂
𝝈𝒎𝒂𝒙 = 𝟏𝟔𝟔. 𝟔𝟕 𝑴𝑷𝒂
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Sample Problem 2.23
The figure shows a roof truss and the detail of the
connection at joint B. Members BC and BE are angle
sections with thickness shown in the figure. The working
stresses are 70MPa for shear in rivet and 140MPa for
bearing stress due to the rivets. How many 19-mm
diameter rivets are required to fasten the said members
to the gusset plate?
D
DETAIL OF JOINT B
14 m m thick
GUSSET PLATE
75X75X13 m m
75X75X6 m m
B
G
4m
P BE
3m
3m
C
A
3m
E
3m
50KN
F
3m
75KN
H
3m
P BC
50KN
Ans.
𝟒 𝒑𝒄𝒔 𝒐𝒇 𝒓𝒊𝒗𝒆𝒕𝒔 𝒇𝒐𝒓 𝒎𝒆𝒎𝒃𝒆𝒓 𝑩𝑪
𝟏 𝒑𝒄 𝒐𝒇 𝒓𝒊𝒗𝒆𝒕 𝒇𝒐𝒓 𝒎𝒆𝒎𝒃𝒆𝒓 𝑩𝑬
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2.4 THIN-WALLED PRESSURE VESSEL
• often cylindrical or spherical (pressure tanks and pipes)
• pressure acting on the inner surface of the cylinder is resisted by
tensile stresses in the walls of the vessel.
!
• Thin-walled if the ratio of ≥ 10.
"
r
t
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2.4 THIN-WALLED PRESSURE VESSEL
cylindrical pressure vessel
spherical pressure vessel
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
a) Circumferential Stress, σc
b) Longitudinal Stress, σl
(a)
cylindrical pressure vessel
(b) FDB for computing the circumferential stress, σc
(c)
FDB for computing the circumferential stress, σl
cylindrical pressure vessel made
of curved sheets
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
a) Circumferential Stress, σc
• normal stress component on a longitudinal section
• also known as Hoop Stress or Tangential Stress
• acts tangent to the surface of cylinder
𝝈𝒄 =
𝒑𝑫
𝟐𝒕
Where:
𝝈𝒄 = 𝒄𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍/𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒊𝒂𝒍/𝒉𝒐𝒐𝒑 𝒔𝒕𝒓𝒆𝒔𝒔
𝒑 = 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒐𝒓 𝒃𝒖𝒓𝒔𝒕𝒊𝒏𝒈 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆
𝑫 = 𝒊𝒏𝒏𝒆𝒓 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝒕 = 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒗𝒆𝒔𝒔𝒆𝒍
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
a) Circumferential Stress, σc
σc
σc
D
L
𝝈𝒄 =
𝒑𝑫
𝟐𝒕
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
a) Circumferential Stress, σc
Sample of actual failure due to circumferential stress
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
b) Longitudinal Stress, σl
• normal stress component on a transverse section
• acts parallel to the longitudinal axis of the cylinder
𝝈𝒍 =
𝒑𝑫
𝟒𝒕
Where:
𝝈𝒍 = 𝒍𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍 𝒔𝒕𝒓𝒆𝒔𝒔
𝒑 = 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒐𝒓 𝒃𝒖𝒓𝒔𝒕𝒊𝒏𝒈 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆
𝑫 = 𝒊𝒏𝒏𝒆𝒓 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝒕 = 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒗𝒆𝒔𝒔𝒆𝒍
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2.4.1 SRESSES DEVELOPED IN THIN-WALLED PRESSURE CYLINDERS
b) Longitudinal Stress, σl
σl
𝝈𝒍 =
𝒑𝑫
𝟒𝒕
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2.4.2 SPHERICAL VESSEL
• same analysis as longitudinal stress in cylinders
• stress is the same regardless of direction
𝝈=
𝒑𝑫
𝟒𝒕
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2.4.2 SPHERICAL VESSEL
P
𝝈=
𝒑𝑫
𝟒𝒕
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Sample Problem 2.24
The cylindrical portion of the propane tank has an outer diameter of 12
in. and a wall thickness of 0.125 in. Calculate the longitudinal and
circumferential stresses (psi) in the wall of the cylinder when the tank is
pressurized to 200 psi.
Ans.
𝝈𝒄 = 𝟗, 𝟒𝟎𝟎 𝒑𝒔𝒊
𝝈𝒍 = 𝟒, 𝟕𝟎𝟎 𝒑𝒔𝒊
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Sample Problem 2.25
Calculate the minimum wall thickness (in) for a cylindrical vessel that is
to carry a gas at a pressure of 1400 psi. The diameter of the vessel is 2 ft,
and the stress is limited to 12 ksi.
Ans.
𝟏. 𝟒 𝒊𝒏
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Sample Problem 2.26
A cylindrical pressure vessel is fabricated from steel plating that has a
thickness of 20 mm. The diameter of the pressure vessel is 450 mm and
its length is 2.0 m. Determine the maximum internal pressure (MPa) that
can be applied if the longitudinal stress is limited to 140 MPa, and the
circumferential stress is limited to 60 MPa.
Ans.
𝟓. 𝟑𝟑 𝑴𝑷𝒂
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Sample Problem 2.27
A spherical shell with 70-in. outer diameter and 67-in. inner diameter
contains helium at a pressure of 1200 psi. Compute the stress (psi) in the
shell.
Ans.
𝟏𝟑, 𝟒𝟎𝟎 𝒑𝒔𝒊
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Sample Problem 2.28
A spherical pressure vessel has a 1.5-ft inner radius and 3/16-in. wall
thickness. If the working tensile stress of the material is 6000 psi,
determine the maximum allowable internal pressure (psi).
Ans.
𝟏𝟐𝟓 𝒑𝒔𝒊
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Sample Problem 2.29
A cylindrical steel pressure vessel 400 mm in diameter with a wall
thickness of 20 mm, is subjected to an internal pressure of 4.5 MN/m2.
a) calculate the tangential and longitudinal stresses in the steel vessel.
b) to what value may the internal pressure be increased if the stress in
the steel is limited to 120 MN/m2?
Ans.
a) 𝝈𝒄 = 𝟒𝟓 𝐌𝐏𝐚; 𝝈𝒍 = 𝟐𝟐. 𝟓 𝐌𝐏𝐚
b) increased by 𝟕. 𝟓 𝐌𝐏𝐚
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Sample Problem 2.30
The tank shown in the figure is fabricated from 1/8-in steel plate.
Calculate the maximum longitudinal and circumferential stress caused by
an internal pressure of 125 psi.
Ans.
𝝈𝒄 = 𝟐𝟏, 𝟎𝟎𝟎 𝒑𝒔𝒊
𝝈𝒍 = 𝟔, 𝟓𝟔𝟔. 𝟎𝟐 𝒑𝒔𝒊
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