See the other picture for type of loading ▪ Derive the following equation of maximumdeflection using double integration method.MAX8Pb(1²-6²)3²9√31EIPb8= (31²-46²) at the center, if a>b48EImaxat x=√(1²-6²)/3M1²16EI8=-M1²9√3 EIat x =at the center1√√3Ômax =at x =&c=8maxWL4120EIWL³384EIWL³192EI 7.y =y =MAX8=11.AaPb48EI15. WPbx(1²-x²-b²) for 0bR₂Ww = N/m- at x = √(1²-6²)/3www. 88TBXW8.at x =y8 max8.Ômax =max8=.Mlx6EIMI²16EIM1²9√3 EIWL4120EIWL³384EI40,WL³192EIM1¹-72)at the centerat x =C1√√3

As all are separate questions hence as per guidelines solving fist one only.

Answered: ¸ = Pb(1²−6²)³² 9√3 IEI 8=- Pb - at…

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Civil Engineering

¸ = Pb(1²−6²)³² 9√3 IEI 8=- Pb - at x=√(1²-6²)/3 (31² - 46²) at the center, if a > b 48EI

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

See the other picture for type of loading

▪ Derive the following equation of maximum
deflection using double integration method.
MAX
8
Pb(1²-6²)3²
9√31EI
Pb
8= (31²-46²) at the center, if a>b
48EI
max
at x=√(1²-6²)/3
M1²
16EI
8=-
M1²
9√3 EI
at x =
at the center
1
√√3
Ômax =
at x =
&c=
8max
WL4
120EI
WL³
384EI
WL³
192EI

7.
y =
y =
MAX
8=
11.
A
a
Pb
48EI
15. W
Pbx
(1²-x²-b²) for 0<x<a
61EI
Pb
OLET [ { (x− a )² + (1²-b³²)x-x²]
61EI
for a<x<l
I
13.Total W = WL
b
40,
Pb(1²-6²2
9√31EI
(31²-4b²) at the center, if a>b
R₂
W
w = N/m
- at x = √(1²-6²)/3
www. 8
8
TB
X
W
8.
at x =
y
8 max
8.
Ômax =
max
8=.
Mlx
6EI
MI²
16EI
M1²
9√3 EI
WL4
120EI
WL³
384EI
40,
WL³
192EI
M
1
¹-72)
at the center
at x =
C
1
√√3

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