1. CTC / MTC 222 Strength of Materials
Chapter 10
Combined Stresses
Principal Stresses

2. Basic Stresses Direct tension: σ = F/A Direct compression: σ = -F/A
Maximum stress due to bending: σmax = Mc / I = M/S
Maximum stress occurs at outermost fiber
For +M, stress is compression on the top, tension on the bottom
Stress at any point: σ = My / I

3. Basic Stresses Direct shear:  = F / As Torsional shear: max = Tc / J
Stress is uniform over area As
Torsional shear: max = Tc / J
Stress at any point:  = Tr / J
Shear stress in beams:  = VQ / It
Stress varies over cross-section
Maximum stress occurs at N.A, unless cross-section is thinner at some point

4. Combined Normal Stresses
Combined stresses due to tension plus bending or compression plus bending can be calculated by a simple algebraic sum of the stresses
This is called superposition
Example: A beam or column subject to combined compression and bending
On one side of the member, σ = F/A + M/S
On the other side, σ = F/A - M/S
Calculation of combined stress due to tension plus bending is similar

5. Maximum Shear Stress Theory of Failure
When stress due to bending, σ, occurs at the same location as shear stress, , the two types of stress combine to produce a larger shearing stress, max
max = √[(σ / 2)2 + 2]
Maximum Shear Stress Theory of Failure
When max > Sys , failure will occur
Shows good correlation with test results for ductile materials

6. Principal Stresses – Initial Stress Element
To analyze a state of combined stress, an initial stress element with a known alignment or orientation is chosen
The normal stresses σx and σY and the shear stresses xy and yx acting on the stress element can be calculated using the formulas for basic stresses
This initial stress element can be used to find the maximum principal stresses and maximum shear stress at that point

7. Principal Stresses
If the initial stress element is rotated, the normal stresses and shear stresses will change
We are interested in finding the orientation where the normal stress is maximum or the orientation where the shear stress is maximum
The angle of rotation, Φ, where the normal stresses are at a maximum on two opposite faces and at a minimum on the two adjacent faces can be found
These normal stresses σ1 and σ2 are called the principal stresses
σ1 – the maximum principal stress
σ2 – the minimum principal stress

8. Principal Stresses
The planes on which the principal stresses act are called the principal planes
On the element on which the principal stresses act, the shear stress is zero
Angle Φ locating the principal planes can be calculated
Φ = tan-1 [ - xy / ½ (σx – σY)]
Principal stresses can also be calculated
σ1 = ½ (σx +σY) + √[[(σx – σY) 2] + xy2 ]
σ2= ½ (σx +σY) - √[[(σx – σY) 2] + xy2 ]

9. Maximum Shear Stress Angle Φ’ locating the maximum shear stress
Φ’ = tan-1 [½ (σx – σY) / xy]
This is 45° from plane of principal stresses
Maximum shear stress
max = √[[(σx – σY) 2] + xy2 ]
Normal stress on maximum shear stress element
σ1= σ2 = σAVG = ½ (σx +σY)

10. Mohr’s Circle for Stress
A graphical aid for finding the stress at any location in a stress element
Constructing Mohr’s Circle is a 15 step process
See examples in textbook
Link to animated illustration of Mohr’s Circle