1. Simple Stress
1.1 Introduction
Statics of rigid bodies & dynamics of rigid bodies are concerned with bodies for which the deformation is
neglected.
Mechanics of deformable bodies deal with the relations between externally applied loads and their
internal effects on bodies. In here, the deformations although small are of major interest.
The difference between the mechanics of rigid bodies and strength of materials are emphasized by
considering the figure shown below.
In statics, the force is to be determined to support the load W. It can be solved by summing up moments
about the pin or hinge equal to zero. In strength of materials, the bar is to be investigated whether it will
break or bend.
1.2 Analysis of Internal Forces
Consider a body of any shape that is acted upon by external forces as shown in the figure.
Figure 1.2.1 Arbitrary shape of a body with the exploratory section.
An investigation of the internal distribution of forces can be done by passing an exploratory
section “a – a” through the body.
The internal forces acting on the exploratory section are necessary to maintain the equilibrium of
either segment as shown in Figure 1.2.2.
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Figure 1.2.2 Distribution of the internal effect on exploratory section a – a.
Pxx – Axial force (It measures the pulling or pushing action perpendicular to the section.)
Pxy , Pxz – Shear forces (These forces measure the resistance to sliding the portion to one side of
the exploratory section past the other.)
Mxx – Torque ( It measures the resistance to twisting the member.)
Mxy , Mxz – Bending Moments (These moments measure the resistance to bending the member
about the y or z axes.)
Stress of a material,  is given by the equation,
=
𝑃
𝐴
𝑤ℎ𝑒𝑟𝑒: 𝜎 = 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛
𝑁
𝑚2
𝑃 = 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑙𝑜𝑎𝑑 𝑖𝑛 𝑁
𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑖𝑛 𝑚2
In SI system of units, the force is in newtons (N), the area is in square meters (m2), the stress is in
newtons per square meter (N/m2). 1 N/m2 = 1 Pa (pascal).
In U.S. Customary Units, the force is in pounds (lb), the area is in square inches (in. 2), the stress is
in pounds per square inch (lb / in.2), commonly in psi.
The above equation for stress is an average stress. The exact expression of stress is obtained by
dividing the differential load dP by the differential area dA. In equation form,
=
𝑑𝑃
𝑑𝐴
A simple stress is a condition under which the stress is constant or uniform.
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As an example, consider a cutting plane that isolates the lower half on a vertical cylindrical bar
(refer to Figure 1.2.3).
Figure 1.2.3 Free-body diagram of the
lower half portion of cylindrical bar.
∑ 𝐹𝑧 = 0
𝑃 = ∫ 𝑑𝑃
𝑃 = ∫ 𝜎𝑑𝐴
∑ 𝑀𝑦 = 0
𝑃𝑏 = ∫ 𝑥𝑑𝑃
𝑃 = ∫ 𝑥(𝜎𝑑𝐴)
If the stress distribution is to be constant then,
𝑃 = 𝜎 ∫ 𝑥(𝑑𝐴)
𝑃𝑏 = (𝜎𝐴)𝑏 = 𝜎 ∫ 𝑥(𝑑𝐴)
𝑏=
∫ 𝑥(𝑑𝐴)
= 𝑥̅ in which
𝐴
b is recognized as the x coordinate of the centroid of the section
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Saint Venant’s Principle:
“The difference between the effects of two different but statically equivalent load becomes very
small at sufficiently large distances from the load. “
Figure 1.2.4 Normal stress distribution in a strip caused by a concentrated load.
Figure 1.2.5 Normal stress distribution in a grooved bar.
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Sample Problems:
1. The bar ABCD in the figure 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. AB = 3330 psi (T); BC = 2780 psi (C); CD = 4380 psi (C)]
2. For the truss shown in the figure, determine the stress in members AC and BD. The crosssectional area of each member is 900 mm2. [Ans. AC = 59.3 MPa (T); BD = 74.1 MPa (C)]
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SHEARING STRESS
Shearing stress (tangential stress) is caused by forces acting along or parallel to the area resisting the
forces.
A shearing stress is produced whenever the applied loads cause one section of a body to tend to slide
past its adjacent section.
Examples:
=
𝑉
𝐴𝑠
where:  = 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑃𝑎 𝑜𝑟 𝑁/𝑚2
𝑉 = 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑁
𝐴𝑠 = 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎 𝑖𝑛 𝑚2
Sample Problems:
1. Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm-diameter
pin at B that support the beam in the figure. [Ans. A = 34 MPa; B = 17.7 MPa]
2. If the wood joint in the figure has a width of 150 mm, determine the average shear stress
developed along shear planes a-a and b-b. For each plane, represent the state of stress on an
element of the material. [Ans. a = 200 kPa; B = 160 kPa]
3. The inclined member in the figure is subjected to a compressive force of 600 lb. 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 DB.
[Ans. AB = 240 psi; BC = 160 psi;  = 80 psi]
BEARING STRESS
Bearing stress is a contact pressure exerted between separate bodies.
Examples:
1. Soil pressure beneath column footings
2. Contact pressure between a rivet or bolt and the contact surface of the plate against which it
pushes
Bearing stress is not constant
Source: Strength of Materials by Pytel & Singer, 4 th Edition
𝑏 =
𝑃𝑏
𝐴𝑏
where: 𝑏 = 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑃𝑎 𝑜𝑟 𝑁/𝑚2
𝑃𝑏 = 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑁
𝐴𝑏 = 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎 𝑖𝑛 𝑚2
Sample Problems:
1. The lap joint shown in the figure is fastened by four rivets of ¾-in. diameter. Find the maximum
load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for
bearing in the plate. Assume that the applied load is distributed evenly among the four rivets,
and neglect friction between the plates.
[Ans. 24,700 lb]
Assignment No. 1 (For normal stress, shearing stress, & bearing stress)
Answers:
Prob. 1-31 →  = 119 MPa
1-33 →  =
𝑃𝑠𝑖𝑛2 𝜃
𝐴
1-34 → D = 13.3 MPa
E = 70.7 MPa
1
2
3
4
1-85 → 𝑙𝐴 = 𝑖𝑛. ; 𝑙𝐵 = 𝑖𝑛.
THIN-WALLED PRESSURE VESSEL
Thin wall refers to a vessel having an inner-radius-to wall-thickness ratio of 10 or more (r/t  10).
Cylindrical or spherical vessels are used in industry to serve as boilers or tanks.
Consider a cylindrical tank carrying a fluid (gas or liquid) under a pressure p as shown in Fig. 4.1.
Figure 4.1 Cylindrical tank
.
Figure 4.1a Cross-sectional area of the cylinder.
Figure 4.1b FBD of the lower portion of section “A-A”.
The wall of the cylindrical tank is subjected to tensile stress due to the internal pressure in the fluid. The
tensile stress is developed across the longitudinal section and transverse section.
The tensile stress developed across the longitudinal section is derived by referring to Fig. 4.1b. A
differential force dF is acting normal to an element located at an angle . In equation form,
𝑑𝐹 = 𝑝𝑑𝐴 = 𝑝𝐿
𝐷
𝑑𝜃
2
A not shown similar differential force dF acts on the symmetrically placed element on the other side of
the vertical centerline. The horizontal components of the forces cancel out. The bursting force F is the
summation of the vertical components of those elementary forces.
𝜋
𝐹 = ∫ (𝑝𝐿
0
𝐹 = 𝑝𝐿
𝐷
𝑑𝜃) 𝑠𝑖𝑛𝜃
2
𝐷
°
[−𝑐𝑜𝑠𝜃]180
0
2
𝐹 = 𝑝𝐷𝐿
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A simpler way of find for F is to consider Fig. 4.2.
F
Figure 4.2 Lower portion of cutting plane “A – A”.
𝐹 = 𝑝𝐷𝐿
By summing up forces vertical equal to zero of the FBD above will result to the equation,
2𝑃 = 𝐹
2𝜎𝑡 𝐴 = 𝑝𝐷𝐿
𝜎𝑡 =
𝑝𝐷𝐿
2𝑡𝐿
𝝈𝒕 =
𝒑𝑫
𝟐𝒕
where:
𝜎𝑡 = 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑎𝑐𝑡𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟) 𝑖𝑛
𝑝 = 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛
𝑁
𝑜𝑟𝑃𝑎
𝑚2
𝑁
𝑜𝑟 𝑃𝑎
𝑚2
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑖𝑛 𝑚
𝑡 = 𝑤𝑎𝑙𝑙 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑖𝑛 𝑚
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In deriving the transverse stress, refer to Figure 4.3.
Figure 4.3 Lower portion of cutting plane “A – A”.
The resultant tensile force acting across the wall section is
𝑃 = (𝜋𝐷𝑡)𝜎𝑙
The bursting force, F is acting on the transverse section and it is
𝐹=(
𝜋𝐷 2
)𝑝
4
Summing up forces along the axis of symmetry, P = F.
(𝜋𝐷𝑡)𝜎𝑙 = (
𝝈𝒍 =
𝜋𝐷 2
)𝑝
4
𝒑𝑫
𝟒𝒕
where:
𝜎𝑙 = 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑎𝑐𝑡𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑡𝑎𝑛𝑘 ) 𝑖𝑛
𝑝 = 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛
𝑁
𝑜𝑟𝑃𝑎
𝑚2
𝑁
𝑜𝑟 𝑃𝑎
𝑚2
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑖𝑛 𝑚
𝑡 = 𝑤𝑎𝑙𝑙 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑖𝑛 𝑚
For thin-walled spherical pressure vessel (Figure 4.4),
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Figure 4.4 Sections of spherical pressure vessel
By summing up forces vertical equal to zero,
𝐹=𝑃
𝑝
𝜋𝐷 2
= 𝜎𝜋𝐷𝑡
4
𝝈=
𝒑𝑫
𝟒𝒕
Example:
A large pipe, called a penstock in hydraulic work is 1.5 m in diameter. Here it is composed of wooden
staves bound together by steel hoops, each 300 mm2 in cross-sectional area, and is used to conduct
water from a reservoir to a powerhouse. If the maximum tensile stress permitted in the hoops is 130
MPa, what is the maximum spacing between hoops under a head of water of 30 m? (The mass density of
water is 1000 kg/m3.)
[Ans. L = 177 mm]
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Assignment No. 2
1. A cylindrical pressure vessel of steel material is of 400 mm in diameter with a wall thickness of
20 mm, is subjected to an internal pressure of 4.5 MPa.
(a) Calculate the tangential and longitudinal stress in the steel.
(b) To what value may the internal pressure be increased if the stress in the steel is limited to
120 MPa?
(c) If the internal pressure were increased until the vessel burst, sketch the type of fracture that
would occur.
[Ans. (a) 45.0 MPa, 22.5 MPa ; (b) 12.0 MPa]
2. The wall thickness of a 1.2-m-diameter spherical tank is 8 mm. Calculate the allowable internal
pressure if the stress is limited to 55.16 MPa.
[Ans. 1.47 MPa]
3. The shown in the Figure P-141 is fabricated from 1/8-in. steel plate. Calculate (a) the maximum
longitudinal and (b) circumferential stress caused by an internal pressure of 125 psi.
[Ans. (a) Max. l = 6,566 psi; (b) Max. t
= 21,000 psi]
4. A cylindrical pressure vessel has an inner diameter of 4 ft and a thickness of ½ in. Determine the
maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal
stress component exceeds 20 ksi. Under the same conditions, what is the maximum internal
pressure that a similar-size spherical vessel can sustain?
[Ans. 417 psi ; 833 psi]
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