Question

LESSUI 9.3 Problem 9.19 Part D Determine the equivalent state of stress on an element at the same point which represents the principal stress, and the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown in the figure below. Suppose that T = 24 MPa (Figure 1) Determine the shear stress Try that acts on the element with orientation 0 .12.8° Express your answer as an integer. IP Ad vec Ro? - 106 MPa MPa Figure Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining 100 MPa Part E Determine the orientation of the planes of maximum in plano shear stress Express your answers using three significant figures separated by a comma. V AXP vec O ?

Answer 1

Given: 0x = - 100 MPa; Oy = 0 ksi Txy = 24 MPa Principal stresses gilon - Oxtoy J 793) * + Fod - -100+0+ JC-100-)* + (24) -100 +0 2 V = -50 + 55.462 01 = -50 + 55.462 = Oz = -50 - 55.462 = 5.462 MPa - 105.462 MPa Orientation of Principal plane ταν tan 20 = (ox- Oy)/2 24 (-100 - 0)/2 = -0.48 tan-1 (-0.48) = -12.82°

Oy - COS 20 + Try sin 20 2 (x *03)+*704) cos 20 + Try sin 20 (-100 +)+(-100-9) cos 21-12.829 + (24) sin 24-12.82%) -10 -100 - 0 (-12.829) + (24) sin 2(-12.82°) - 50 – 45.077 – 10.385 = - 105.462 MPa = 01 Hence (@p)1 = -12.82° and (Op)2 = -12.82 + 90° = 77.18° Max. in-plane shear stress Tmax-in plane - 10 • 6-10 - 0)*+ (24) - 55.462 MPa 55.462 MPa

Avg. Normal stress σανg Pava - -100 + 0 = Oxfoy - -100 +0 - = - 50 MPa Orientation of Maximum in-plane shear stress tan 260.), - -(0x-Oy)/2 = -(-100 - 0)/2 = = 24 2.083 - Txy (03), = tan-1 (2.083) = 32.180 (03)2 = 32.18 + 90º = 122.18°