Answer to Question #213049 in Mechanical Engineering for Evans

Question

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?

Answer 1

Part a

Part b

$ΣF_x = 0: H_A = 0\\ ΣF_y = 0: - (U_1^{left} *7)0.5- P_1 + R_A = 0\\ ΣM_A = 0: (U_1^{left} *7*0.5) * (9 - 8 + (2/3)*7) + 5*P_1 + M_A = 0;$

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

Part c

Part d

Part e

$Q(x_2) = - ([(U_1^{left} - U_1^{left} *(8 - x)/7)*(x - 1)]/2 + U_1^{left} *(8 - x)/7*(x - 1))\\ Q2(1) = - ([(30 - 30*(8 - 1)/7)*(1 - 1)]/2 + 30*(8 - 1)/7*(1 - 1)) = 0 (kN)\\ Q2(4) = - ([(30 - 30*(8 - 4)/7)*(4 - 1)]/2 + 30*(8 - 4)/7*(4 - 1)) = -70.71 (kN)$

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

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