written 6.5 years ago by | • modified 3.2 years ago |
Subject: Structure Analysis 2
Topic: Determination of Collaps Load for Beams
Difficulty: Hard
written 6.5 years ago by | • modified 3.2 years ago |
Subject: Structure Analysis 2
Topic: Determination of Collaps Load for Beams
Difficulty: Hard
written 6.5 years ago by |
1) Collapse of the Span AP :
With plastic hinges at A,B and at the centre of the Span.
Using Virtual work principle
EWD = IWD
MP=4006=66.67KN.m
MP=66.67Kn.m
3) Collapse of the span CD:
with plastic hinges at C,D and at the centre of the Span.
Using virtual work done:
FWD = IWD
100∗γ=1.5MPθ+1.5MPθ+1.5MPθ+MPθ
θ=γ3γ=3θ
100∗3θ=MPθ(1.5+1.5+1.5+1)
300θ=5.5MPθ
MP=3005.5=54.55KN.m
MP=54.55KN.m
(50∗γ)+(80∗γ)+(50∗γ)=2MPθ+2MPθ+2MPθ+1.5MPθ
MP=5207.5=69.33
MP=69.33Kn,m
2) Collapse of the span BC
with the plastic hinges at B,C and at the centre of the span.
using virtual work done
EWD = IWD
25∗12∗8∗γ=1.5MPθ+1.5MPθ+1.5MPθ+1.5MPθ
θ=γ4γ=4θ
4) Collapse of the span CD
with the plastic hinges at D and understand the load at c.
EWD = IWD
(45∗γ)+(45∗γ)=MPθ+MPθ+MPα
270θ=MPθ(1+1+2)
MP=2704=65.7Kn.m
MP=65.7KN.m
From all the above possible consideration, the collapse Mechanism for the span AB requires the greatest value of MP
MP=69.33KN.m