Question

An arm ABC lying in a horizontal plane and supported at A (see figure below) is made of two identical solid steel bars AB and BC of length L welded together at a right angle. The arm is subjected to its own weight only. Part 1 - Determine the stress state at the top of the bar at the support A, then draw stress distributions for the section at A. (3 points for the drawings) Part 2- Draw Mohr's circle (2 points) for the stress element at the top of section A. Then locate on the circle the principle normal and shear stresses and the angles of the principle planes. Given Self weight of steel = 63 kN/m3 L = 0.35 m Diameter of the bar = 9 mm 4 < Insert you Answers in the following boxes:

Answer 1

Due to bar Bc lood on ab are One trangverse sheai load due to self- weight of BC & moment due to that self-weight- FIL P= 63 x 1 x 0.00880.350 P= 1.402x10-3KW = 7.402 N ем m = moment: Belf weight gacxş 2:1402 * 0,35 Caits like Twisting moment for AB) Now we need to find Bending moment at point a = M = 0.245 Nm MA= P x2 + Self-weight of ABY Ž MA= 1.402x2 + 1 402 x 2 - 4 7.402 x 3 = 10402xŽ x0-35 MAE 01736 Nom moment at A & MA= 0.736 dom Bending moment 4 Twisting TA = 0.245 Airm Top fibsecuta Sexy & Je=0 mpa Part 1 32MA oy: Trds 3 10.283 mla 16TA Tuy = Tye Trd Part 2: Toyons lagi tayo 5.141 + 5.4187 = 1.71 MPa og 10.559 mila, (o) -0.277 MPa (02) Tmax for = 5:418 mla ; R = fryt tay = 10.42 mpa Op Tan 2 -18.4 71.5934 try cici