please solve using DOUBLE INTEGRATION METHOD Derive at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different pointsof the beam, maximum deflection, and slopes as shown in the table.L-(3L² – 4x²) 0 ba

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Answered: Px (3L – x) PL 3EI PL? 2EI 6EI

Engineering

Civil Engineering

Px (3L – x) PL 3EI PL? 2EI 6EI

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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Concept explainers

Beams

In structural engineering, a beam is a component made of a variety of materials (including steel alloys, and wood) that can resist loads applied laterally to the beam axis. Members, elements, rafters, shafts, and purlins are all terms used to describe beams.

Question

please solve using DOUBLE INTEGRATION METHOD

Derive at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different points
of the beam, maximum deflection, and slopes as shown in the table.
L
-(3L² – 4x²) 0 <x<
PL3
Smax =
48EI
PL?
04 = Og =
16EI
Px
Px2
(3L -
6EI
PL3
OB =
3EI
PL²
OB =
2EI
A
x)
48EI
L
Pb(L² – b²)³/²
9/3LEI
L² – b²
Pbx
-(L² – x – b²)
6LEI
Smax
Px2
(За — х) 0<x<a
6 ΕΙ
0<xsa
Pab
(2L a)
6L EI
Ра?
(3L – a)
6EI
Pa?
OB
2EI
8 =
Pb
(x – a)³ + (L² – b²)x –x
at x =
Pab
(2L – b)
6L EI
3
Pa2
(3х — а) а <х<L
6EI
OB =
a
6LEI
Pb
(3L² – 4b²) if a >b
a<x<L
Scœnter =
48EI
Wo
Wo
wox?
(6L² – 4Lx+ x²)
24EI
woL3
OB =
6ΕΙ
B
5woL4
384EI
A
Wox
dB =
8EI
24ET (L – 2Lx²+x³)
Smax
O4 = OB =
24EI
Wo
Wo
Wo
wox?
(6a² – 4ax+x²) 0<xsa
24EI
[a²(a – 2L)² +2a(a – 2L)x² + Lx³] dænter =
24L EI
(5L* – 12L?b² + 86*)
Woa?
(a – 2L)²
384 EI
A
24L EI
Woa3
6ΕΙ
0<xsa
A
woa3
8 = .
if a 2 b
8 =
OB =
(4L — а)| 0в
Woa?
(-L+x)(a² – 4Lx+2x²)
Woa?
-(2L² – a²)
OB =
24L EI
24EI
woa*
(4х — а)
24EI
Woa?
(3L² – 2a²) if asb
96EI
a <x<L
24L EI
dcenter =:
L
a<x<L
Wo
wo
Smax = 0.006 522-
EI
7woL3
Wox
360EI
Wox?
: (10L³ – 10L²X+ 5Lx² – x³) | dB =
(7L* – 10L²×² + 3x*)
B
A
A
at x = 0.5193L
OB =
24 EI
360L EI
120L EI
30EI
Sœnter = 0.006 510-
EI
OB =
45 EI
Mo
M,L²
8 max
9/3EI
M,L
Mox?
8 =
M,L²
SB =
2EI
MọL
OB
at x=
V3
%3D
A
B
Mox
6EI
A
(L² – x²)
6L EI
2EI
EI
M,L?
16EI
MọL
OB =
3EI
Mo
dæenter =

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