Question

A plastic bar fits snugly between rigid supports at room temperature (68°F) but with no initial stress (see figure). When the temperature of the bar is raised to 160°F, the compressive stress on an inclined plane pq becomes 700 psi. What is the shear stress on plane pq? (Assume a = 30 x 10-6°F and E = 450 x 10 psi.) Figure 4

Answer 1

450 x 10² psi X 30x 10" Initial temperator - 68°F 20°c temperature 160°F final 7111°C due to rise in temperature Compressive Stress develop ny ܠܠܠܓܓܠ 9 EDIT %** 92°F lau2000 x103 psi 1242 psi (compressive) so it must tokeo as (-xe) Ox = -1942 psi principal plona exy centre of mohlo circle cou = Othey = 621551 P Radius = al 2 7242725 20 (2) 70 -780 - Geri of 621Psi

on p q plane Compressive stress is = 7oopsi Rcosa =(700-621) 621 Cosa = (100-621) 6al cosa = 79 cosa = 0.1972 d=82.69° shear stress = R sind = 621. Sin 82.690 sheaostress at pa = 618-9 psi
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