Derive at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different points Phx -- of the beam, maximum deflection, and slopes as shown in the table. PL Px 48EI (M - 4x²) 0sxs PL? 16E1 Px² PL 6EI (3L – x) PL² 2EI 3EI Phx (1? – x² – b²) Px2 6LEI 9/3LEI GET Sa - x) 0sxsa Pab (2L-a) 6L. EI 6EI Pa Pa đn =7 (3L - a) Ph (L. LEI -a)' + (L² –)x -x at xV3 Pab (21.-b) 6LEI 6EI 2EI Pa :(3x-a) as xSL 6EI asxSL daner (3M - 46) if a >b 48EI SEI Swl" 384EI 24E7 (61² – 4Lx +x²) (L' – 21.x² + x') 6EI 24EI 24EI (52* – 12136² + 86*)| if a zb (31² – 2a") if a sb Wo 24L ET"(a- 21.) + 2a(a -- 21.)x² + Lx'|d 384EI 241. EI 2I (6a - 4ax + x²) 0sxsa "2AL E la- 21.)* 241. EI 24EI Woa (4L - a) O= 24EI 6EI (21 -a) 24LEI 241, ET-L+ x)(a² – 4Lx + 2x²) Santer (4x- a) asxSL 241. EI 24EI asxsL 96E7 dun -0.006 522 7wl 360EI 360L ET (7L* – 10L²r²+3x*) at x-0.5193L 360LEI 1 20L EI 120L. ET (10L' – 10L²X+ 5Lx² – x³) ] ón 30EI 24EI daner-0.006 510- El 45EI Mol. Mox? at x Max (L² - x²) 61.EI 6EI O EI 2EI 2EI Mol. Mo Saner 16EI 3EI

Derive at least 5 cases from the 12 cases of beam loadings that is shown. Derive the deflection at different points Phx -- of the beam, maximum deflection, and slopes as shown in the table. PL Px 48EI (M - 4x²) 0sxs PL? 16E1 Px² PL 6EI (3L – x) PL² 2EI 3EI Phx (1? – x² – b²) Px2 6LEI 9/3LEI GET Sa - x) 0sxsa Pab (2L-a) 6L. EI 6EI Pa Pa đn =7 (3L - a) Ph (L. LEI -a)' + (L² –)x -x at xV3 Pab (21.-b) 6LEI 6EI 2EI Pa :(3x-a) as xSL 6EI asxSL daner (3M - 46) if a >b 48EI SEI Swl" 384EI 24E7 (61² – 4Lx +x²) (L' – 21.x² + x') 6EI 24EI 24EI (52* – 12136² + 86)| if a zb (31² – 2a") if a sb Wo 24L ET"(a- 21.) + 2a(a -- 21.)x² + Lx'|d 384EI 241. EI 2I (6a - 4ax + x²) 0sxsa "2AL E la- 21.) 241. EI 24EI Woa (4L - a) O= 24EI 6EI (21 -a) 24LEI 241, ET-L+ x)(a² – 4Lx + 2x²) Santer (4x- a) asxSL 241. EI 24EI asxsL 96E7 dun -0.006 522 7wl 360EI 360L ET (7L* – 10L²r²+3x*) at x-0.5193L 360LEI 1 20L EI 120L. ET (10L' – 10L²X+ 5Lx² – x³) ] ón 30EI 24EI daner-0.006 510- El 45EI Mol. Mox? at x Max (L² - x²) 61.EI 6EI O EI 2EI 2EI Mol. Mo Saner 16EI 3EI

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

please solve using MOMENT AREA 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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