Answer to Question #213049 in Mechanical Engineering for Evans

Question #213049

The beam shown in Figure P2 is a simply supported beam with a rectangular

cross-section.

A. a. Draw the free body diagram for the beam.

B. b. Calculate all the reaction forces at the supports.

C. c. Draw the shearing force diagram for the beam on a graph sheet.

D. d. Draw the bending moment diagram for the beam on a graph sheet.

E. e. What is the maximum normal stress in the beam?


1
Expert's answer
2021-07-07T06:43:01-0400

Part a



Part b

"\u03a3F_x = 0: H_A = 0\\\\\n\u03a3F_y = 0: - (U_1^{left} *7)0.5- P_1 + R_A = 0\\\\\n\u03a3M_A = 0: (U_1^{left} *7*0.5) * (9 - 8 + (2\/3)*7) + 5*P_1 + M_A = 0;"

"H_A = 0 (kN)\\\\\nRA = (U_1 ^{left} *7)\/2 + P_1 = (30*7)*0.5 + 40 = 145.00 (kN)\\\\\nMA = - (U_1^{left} *7\/2) * (9 - 8 + (2\/3)*7) - 5*P_1 = - (U_1^{left} *7\/2) * (9 - 8 + (2\/3)*7) - 5*40 = -795.00 (kN*m)"

Part c



Part d



Part e

"Q(x_2) = - ([(U_1^{left} - U_1^{left} *(8 - x)\/7)*(x - 1)]\/2 + U_1^{left} *(8 - x)\/7*(x - 1))\\\\\nQ2(1) = - ([(30 - 30*(8 - 1)\/7)*(1 - 1)]\/2 + 30*(8 - 1)\/7*(1 - 1)) = 0 (kN)\\\\\nQ2(4) = - ([(30 - 30*(8 - 4)\/7)*(4 - 1)]\/2 + 30*(8 - 4)\/7*(4 - 1)) = -70.71 (kN)"


"Q_3(4) = - ([(30 - 30*(8 - 4)\/7)*(4 - 1)]\/2 + 30*(8 - 4)\/7*(4 - 1)) - 40 = -110.71 (kN)\\\\\nQ_3(8) = - ([(30 - 30*(8 - 8)\/7)*(8 - 1)]\/2 + 30*(8 - 8)\/7*(8 - 1)) - 40 = -145 (kN)"


"Q_4(8) = - ([(30 - 30*(8 - 8)\/7)*(8 - 1)]\/2 + 30*(8 - 8)\/7*(8 - 1)) - 40 = -145 (kN)\\\\\nQ_4(9) = - (-30*7)\/2 - 40 = -145 (kN)"


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