CTC / MTC 222 Strength of Materials

1. CTC / MTC 222Strength 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 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 • http://web.utk.edu/~wratchuk/mohr/animation.html