0

745views

MODULE 04 - Q.4

**1 Answer**

0

745views

MODULE 04 - Q.4

0

1views

written 5.6 years ago by | • modified 5.6 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:

ADD COMMENT
EDIT

Please log in to add an answer.