ME 206 – Strength of Materials
1. Introduction - Concept of Stress
Course objective: Provide future engineers means of analyzing and designing various machine
parts and load-bearing structures.
Stress: The force per unit area or intensity of the forces distributed over a given section is called
stress on that section.
𝜎=
𝑃
𝐴
𝑃: 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝐴: 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
Sign convention:
(+) Tensile
(−) Compression
Units:
F N
=
= Pa
A m2
𝝈
N
= MPa
mm 2
𝑨
𝝈=
𝑷
𝑨
1 GPa = 103 MPa = 106 kPa = 109 Pa
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Type of problems:
P (load) and A (area) given → determine σ and choose appropriate material with 𝜎 < 𝜎𝑦𝑖𝑒𝑙𝑑
σ (Stress) and A (area) given → determine the load bearing capacity.
σ (Stress) and P (load) given → determine the size of the cross-section (A).
Tensile Stress
Consider a bar of uniform cross-section that is loaded in tension.
If we take a section of this bar, there will be an internal load P to provide the static balance. This
internal load will not be a point load, it will be a distribution over the cross section, resulting in
stress over the cross section.
When the internal load is perpendicular to the cross-section, then stress over this cross-section is
called normal stress.
In reality, stress can vary with position. Therefore, we can write:
𝑃 = ∫ 𝑑𝐹 = ∫ 𝜎𝑑𝐴
𝐴
𝐴
The stress variation is more significant next to a point load. As you move away from the point
load, stress distribution becomes more uniform and gets closer to the average value.
In this course, in these type of loading conditions, we will assume uniform stress distribution for
simplicity.
𝜎 = 𝜎𝑎𝑣𝑒 =
𝑃
𝐴
In real applications, be careful when analyzing regions close to a concentrated load.
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𝒅 𝑭 = σ dA
𝒅𝑨
Note that this uniform stress assumption is valid when the line of action of the concentrated load
passes through the centroid (geometric center) of the section. In other words, there should not be
any bending moment over the cross section that is analyzed.
Consider the below example:
From the equilibrium of A⸍B, the internal forces at section B
are:
A
M=Pd
force P applied at the centroid of the cross-section
moment M of magnitude M = Pd.
B
A⸍
A⸍
The stress distribution has to be equivalent to P and M
combined, therefore the stress can no longer be uniform. We
will learn how to get the stress distribution in such cases later
in the semester.
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Shear Stress
When transverse forces P and P⸍ are applied to a member AB as shown above, the internal force
P is parallel to the cross-section. This type of stress is called shear stress.
Shear force divided by the area gives the average shear stress:
𝜏𝑎𝑣𝑒 =
𝑃
𝐴
Similar to the above discussion for tensile stress, in reality, τ is not uniform. τ is zero at the
surface and it exceeds τave at the center. In this course, we will focus on τave for learning
purposes.
Shear stresses are commonly found in bolts, pins and rivets used to connect various structural
members.
Bearing stress
Bearing stress is the type of stress that occurs as a result of contact pressure. Bolts, pins, rivets
and similar members create these type of stress on the bearing surface of the member in contact.
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𝜎𝑏 =
𝑃
𝑃
=
𝐴 𝑡×𝑑
𝐴 = 𝑡 × 𝑑: Projected area of the rivet hole
Stress on an oblique plane under axial loading
In two-force members, axial forces create normal stress whereas in pins and rivets, the transverse
forces lead to shear stress. Over an oblique cross-section, both stresses might exist at the same
time.
Consider the above two-force member subjected to axial forces P. Let’s take a section forming
an angle θ with respect to the normal plane and draw the free body diagram of the left part.
→
𝐹 : normal force → 𝜎 =
𝐴0
𝐴0
= 𝑐𝑜𝑠 𝜃 → 𝐴𝜃 =
𝐴𝜃
𝑐𝑜𝑠 𝜃
𝑉
= 𝑠𝑖𝑛 𝜃 → 𝑉 = 𝑃 𝑠𝑖𝑛 𝜃
𝑃
Therefore,
𝜎=
𝐹
𝐴𝜃
→
𝑉 : shear force → 𝜏𝑎𝑣𝑒 =
𝐹
= 𝑐𝑜𝑠 𝜃 → 𝐹 = 𝑃 𝑐𝑜𝑠 𝜃
𝑃
𝐹
𝑃 𝑐𝑜𝑠 𝜃
𝑃
=
=
𝑐𝑜𝑠 2 𝜃
𝐴0
𝐴𝜃
𝐴0
𝑐𝑜𝑠 𝜃
𝜏𝑎𝑣𝑒 =
𝑉
𝑃 𝑠𝑖𝑛 𝜃
𝑃
=
=
𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜃
𝐴0
𝐴𝜃
𝐴0
𝑐𝑜𝑠 𝜃
For the special cases of 0, 45 and 90 degrees:
𝜃 = 0° →
𝜎 = 𝜎𝑚𝑎𝑥 = 𝑃/𝐴0
𝜃 = 90° → 𝜎 = 0
𝜃 = ±45° → 𝜎 = 𝑃/2𝐴0
𝜏𝑎𝑣𝑒 = 0
𝜏𝑎𝑣𝑒 = 0
𝜏𝑎𝑣𝑒 = 𝑃/2𝐴0
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𝑉
𝐴𝜃
Stress under general loading conditions: Components of stresses
Consider a body subjected to several loads/forces to determine the
stress distribution at point Q within the body.
First pass a plane parallel to yz-plane that contains point Q.
⃗⃗⃗⃗𝑥 acts perpendicular to the plane containing ∆𝐴
𝛥𝐹
⃗⃗⃗⃗⃗𝑥 acts parallel to the plane containing ∆𝐴.
𝛥𝑉
𝑥
⃗⃗⃗⃗⃗𝑥
⃗⃗⃗⃗⃗
𝛥𝑉
𝑦 is the component of 𝛥𝑉 along y axis.
⃗⃗⃗⃗⃗𝑥 along z axis.
⃗⃗⃗⃗⃗𝑧𝑥 is the component of 𝛥𝑉
𝛥𝑉
Now, we can write the three stress components generated by these forces:
F x
 x = lim
A→ 0 A
 xy = lim
A → 0
 xz
V y x
A
Vz x
= lim
A→ 0 A
In subscript xy: x is denotes the normal vector of the surface (first subscript) and y indicates the
direction of the stress component (second subscript).
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Sign Convention
Consider a small cube centered at Q with stresses exerted on each face of the cube.
Now, cut a section at Q parallel to the xy-plane.
From moment equilibrium ∑ 𝑀𝑧 = 0, we find 𝜏𝑥𝑦 ∆𝐴 − 𝜏𝑦𝑥 ∆𝐴 = 0
→ 𝜏𝑥𝑦 = 𝜏𝑦𝑥
Similarly, 𝜏𝑦𝑧 = 𝜏𝑧𝑦 and 𝜏𝑥𝑧 = 𝜏𝑧𝑥
Therefore, to define the stress state at a point within a member, we need 6 components.
𝜎𝑥 , 𝜎𝑦 , 𝜎𝑧 : Normal Stresses
𝜏𝑥𝑦 , 𝜏𝑥𝑧 ,𝜏𝑦𝑧 : Shear Stresses
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