¸ = Pb(1²−6²)³² 9√3 IEI 8=- Pb - at x=√(1²-6²)/3 (31² - 46²) at the center, if a > b 48EI
¸ = Pb(1²−6²)³² 9√3 IEI 8=- Pb - at x=√(1²-6²)/3 (31² - 46²) at the center, if a > b 48EI
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:▪ 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3df4203b-6d3e-4d3d-b2db-c99c20c87d89%2Fad2c4d6c-ed5d-4aa1-b372-0daacd6c8036%2Funlq0pn_processed.png&w=3840&q=75)
Transcribed Image Text: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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