Consider the truss structure shown below, which is composed of three members of

Question

Consider the truss structure shown below, which is composed of three members of different areas. This truss is loaded with a rightward facing horizontal point load at node C. Member AB, Area 1.0 inA2 Member BC, Area 2.2 in 2 Member AC, Area = 3.0 in^2 20' 15' 5' a) What is the maximum amplitude of the point load at node C such that stress in each truss member remains under 40 ksi? Hints: The structure is linear, so the principle of superposition applies (of our interest in this case, f(ax)-af(x)). b) At the maximum load amplitude found in part A, what is the stress in each member? Hint: One of the stresses should be 40 ksi.

Explanation / Answer

a) let the force be P

let force in members ab, ac and member be cb be Fab,Fac and Fcb repectively

let us consider moment equilibrium at support b

P*20+RA*20=0

RA = -P kips = P kips(downwards)

vertical reaction at suppport B = P kips(upwards)

horizontal reaction at B = P kips (towards left)

Angle of ac with horizontal = tan-1(20/15)=53.13o

Angle of bc with horizontal = tan-1(20/5)=75.96o

Considering joint equilibrium of a:

Facsin53.13-P=0

Fac = 1.25P(tension)

Fab = 1.25Pcos53.13 = 0.75P(compressive)

from equilibrium of joint b:

Fbcsin75.96+P=0

Fbc = 1.03P (compressive)

Stress in ab = 0.75P/1=0.75P ksi

Stress in ac = 1.25P/3=0.417P ksi

Stress in bc = 1.03P/2.2 = 0.468P ksi

the crtical load will be governed by 0.75P

0.75P=40

P=40/0.75=53.33 kips

b)Stress in ab = 0.75*53.33=40 ksi (compression)

Stress in ac = 0.417*53.33=22.22ksi (tension)

Stress in bc = 0.468*53.33=24.96 ksi (compression)

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