Question

90 kN/m 165 KN 120 KN.m ܠܠܠܠ 2m 3m 1m 1m А в 2m С DE F

which among the choices almost gives the moment at the fix support

A. 11.08
B. 165
C. 18.88
D. 0

Answer 1

Answer is C - 18.88

First convert the triangular load to a point load

a triangular udl of w/unit length l = a point load of wl/2 acting at 1/3 from the right angle vertex

(98x312) Trangle load 90 km A im 3m

So trianglular load can be replaced with 135kn point load at 1m from right end.

So we have

Triangular load can be converted to pont load gokN/m 135KN Im 2m 3m 135 kW So we have 165kn 120 KN/m (м 20 Im im 3 TREH im 2m F D E A B PRE "HE 'RB- B

Moment about B - 165(1) +0 - 120 + 135(+) BRE (6) + M=0 8=0 Moment about F ERE+8= 255+ M -165(9) + R$(8) – 120-135(4) + R$(2)+M=0 8R3+2RETM# 2145 8R₂ +2RE = 2145-M Moment about E - 16517) + R$(6) – 120 m 135(2) +M=0 -2RE=0 6R6-2Rp = 1545-M 300 also RB + RETRE 4 equations foror varables

If you solve you will get M = 18.88Kn/m

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