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:

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Answer 1

3 Normonal stress at a 54 - Mx ZA ZA = & 2= Tdt 64 do diameter -gmon Spy = ~32My TdB -32% (-0.135) X(9x10373 Pa Syy - 10.27 MPa shear stresses at a yx = My ZA J ZA= J = 29 = adA 32 Exp = Oxy - 16 My ad² бух x = - 1.It MPa Syz co

G - 10.27 MPa Here, Ox' O Gy - 10.27 MPa ext exy 361 = -l. II MPa 2 Tempa) 4 2 (0,1.71) 2 4 6 12 14 (MPa) -2 -4 (10.27 - d.II) Exty - 2 Gary Sany =) - 5.135 MPa MPa R: 1 2N (10.27-01 +(-1.71-4-71) Ra5.2006 MPa

og Opt 90° +R Serg 19 = 10.341 MPa Gz = Cong-R 6 = -0.071 MPa max 16,6al max 2 Tmax 25.206 MPa Now, tan 20p ll dofL 5.135 1.71 Parig Op=9.2ogo Os = 99.2009