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MODULE 04 - Q.4
1 Answer
written 5.7 years ago by | • modified 5.7 years ago |
Solution:
$\sum M_A=0$
$M_A=12\times 5+ 10\times 3 \times \frac{3}{2}=0$
$M_A=15kN.m$
$\sum F_y=0(\uparrow +ve)$
$V_A-10\\times 3=0$
$V_A=30kN$
$\sum F_x=0(\rightarrow +ve)$
$H_A-12=0$
$H_A=12kN$
$BM_B=15-12\times 5=-45kN.m$
(1)Consider part AB:
(2)Consider part BC:
(3)Consider part CD: