The points A B C are fixed what are the static loads on FAD

The points A, B, C are fixed, what are the static loads on F_AD, F_BD, F_BC?

Solution

solution:

1)here @,A,B,C are fixed points of truss and have zero displacement and moment is zero at support and point D is in intersection of link and applied vertical load of 204.5 N upward

2)here coordinate of pointA,B,C and D are

A(38,8),B(50,1),C(55,0) and D(47,10)

3)hence angle of link Ad is

tanm1=y2-y1/x2-x1=10-8/47-38

m1=12.52

4)angle of link BD with vertical

tanm2=x2-x1/y2-y1=50-47/10-1

m2=18.43

5)for link CD

tanm3=55-47/10-0

m3=38.65

6)hence for static equillibrium of system we get

Fx=0

Fadcos12.52-Fbdsin18.43-Fcdsin38.65=0

Fy=0

Fd=204.5 N

Fd+Fadsin12.52+Fbdcos18.43+Fcdcos38.65=0

7)here are three variable hence we need three equation to solve system ,hence

Ma=0

we get that instead of reaction consider direction original assumed forces

Fbdcos18.43*50-Fbdsin18.43*7+Fcdcos38.65*55-Fcdsin38.65*8+204.5*47=0

so we get that

Fbd=-212.54-.839Fcd

8)putting value in above equation we get equation in Fad and Fcd as

.9762Fad-.3593Fcd=-67.193

.2167Fad-.0149Fcd=-2.86

so we get that

Fad=-.4173 N

Fcd=185.87 N

on putting value we get that

Fbd=-212.54-.839Fcd

Fbd=-368.48 N

10)hence forces in member are

Fad=-.4173 N compression

Fcd=185.87 N tension

Fbd=-368.48 N compression