Question

The state of stress at a point is shown on the element. (a) Determine the principal stress at the point. Specify the associated orientation and indicate the results on an element at the point. (b) Determine the maximum in-plane shear stress at the point. Specify the associated orientation and indicate the results on an element at the point. (c) Determine the state of stress on an element at the same point oriented 40° clockwise with respect to the element shown. 15 ksi

Answer 1

+ 25ksi sokei 2 kfi @ Principal stress at a point is given by 9,2= of tay t Jie -30415 + 63=13*+(1232 7,2= -7.5 + 25-5 I= loksi , 2 = -33 kg; tan 20p= Exy - 12 ex ) -3045 Op=-14.03° 18t principal stress angle got Op = 75.96 - 2nd principal stress angle gotop= 75.96€ Ta 18ksj

6 maximum in-plane shear stress is given by Tmax max J230-153*+(1332 -30-1572 Tmax= 25. 5ksi I n = (1+2) = 7.5ksi gotop Optypo. 7450 ex=30ksi imax 15ksi

© State of stress at angle 0 = 4 % in clockwise is giren by : - (0 = - 45) (for (W 7-ve) Normal stress & o out) + 3) 6820 + Txy Sin20 (30315) + (30-15) Cos2 (-40) + 12 Sin 2145) a = -7.5 + 3.907.-11.817 on=-15.42 ksil Shear stress : G agsint@)- 7 = (x ) sinzo - exy cojzo 7 = 430-15) Sin 2645) -12 08/2645)) = 22.150 -2.0837 2= 20.074 ksi at 0=40° cw. State of stress n = -15.42 ksi - 20.074 K8;