how to we solve the 157625/24 EI and 5775/4EI? onSolving equation andCY= 15762524 JandCy: 5775HETAs per conjugate beary Method:~Oc= Shear force at C in conjugate bear=> 0₁ = −577541521443.75FTand Sc= Bending moment at C in conjugate beam6567.7083157625 =⇒ &c=f24 EJЕТ In conjugateBendingbear! -moment at Hinge B = 0(312.5 xFT=> — Co₁ + (Cyr 5) =Sys:) + Cry + 5ly = 1562524 [TMMA O1-3 - (1 - (-)-(+57] (10-²) + (22+57 (1)312.5Cry + (Cyx 15 =195=) - (y + 15 Cy = 15088-54167175 cóET2(2x 10)

First subtract the first equation from the second equation which gives the value of Cy Substituting…

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Engineering

Civil Engineering

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Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

how to we solve the 157625/24 EI and 5775/4EI?

on
Solving equation and
CY= 157625
24 J
and
Cy: 5775
HET
As per conjugate beary Method:~
Oc= Shear force at C in conjugate bear
=> 0₁ = −5775
415
2
1443.75
FT
and Sc= Bending moment at C in conjugate beam
6567.7083
157625 =
⇒ &c=
f
24 EJ
ЕТ

In conjugate
Bending
bear! -
moment at Hinge B = 0
(312.5 x
FT
=> — Co₁ + (Cyr 5) =
Sys
:) + Cry + 5ly = 15625
24 [T
MMA O
1-3 - (1 - (-)-(+57] (10-²) + (22+57 (1)
312.5
Cry + (Cyx 15 =
195
=) - (y + 15 Cy = 15088-54167
175 có
ET
2
(2x 10)

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