Derive at least 5 cases from the 12 cases of beam loadings that is shown. Using DOUBLE INTEGRATION METHOD. Derive the maximum deflection, slopes and deflection at different points of the beam as shown in the table.
Derive at least 5 cases from the 12 cases of beam loadings that is shown. Using DOUBLE INTEGRATION METHOD. Derive the maximum deflection, slopes and deflection at different points of the beam as shown in the table.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Derive at least 5 cases from the 12 cases of beam loadings that is shown. Using DOUBLE INTEGRATION METHOD. Derive the maximum deflection, slopes and deflection at different points of the beam as shown in the table.
![PL³
Smax =
48EI
PL?
O, = OR =
16EI
Px
Px2
(3L — х)
6EI
PL³
dB
PL²
OB =
2EI
B
(3L² – 4x²) 0 <xs
A
48EI
х
3EI
Pb(L² – b²)³/²
9/3LEI
L² - b²
Pbx
:(L² – x² – b²)
0sxsa
Smax =
6L EI
Px2
(За — х) 0<x<a
6EI
Pab
O, =
(2L – a)
6L EI
В
Ра?
Pa?
OB
8 =
Pb [L
(x– a)³ + (L² – b²)x – x³|
6L EI b
A
at x =
8 =
SB =
(3L – a)
3
Pab
Pa?
6 ΕΙ
2EI
OR =
(2L – b)
6L EI
Pb
(3L² – 4b²) if a > b
48EI
b
(3х — а) а <х<L
6EI
a<x<L
Sœnter =
Wo
Wo
woL³
O4 = Og =
24EI
Wo.x
(L³ – 2Lx² +x³)
Smax =
Wo.x?
(6L² – 4Lx+ x²)
B
woL4
24EI
384EI
A
24EI
OB =
6ΕΙ
8EI
woX
-|a²(a – 2L)² +2a(a – 2L)x² + Lx³] | dœnter =
24L EI
384EI
Wo
(5Lª
12L²6² + 8b*)
Woa?
(a – 2L)²
84 =
wo
0<xsa
24L EI
Wox?
(6a² – 4ax+ x²) 0<x<a
if a 2 b
Woa²
Woa?
24L EI
В
(2L² – a²)
woa³
Woa?
(3L² – 2a²) if asb
96EI
24EI
woa?
(-L+x)(a² – 4Lx+ 2x²)
OR =
A
dB
(4L – a)| 0B
denter =
24LEI
24EI
6EI
a<xsL
Woa3
(4х —
24EI
а)
a <x<L
L
Smax = 0.006 522
EI
7woL3
Wox
(7L* – 10L²x² +3x*)
360L EI
360EI
wo
at x = 0.5193L
woL³
OB =
+7°M
EI
OR =
45 EI
woL4
denter = 0.006 510-
B
Wox-
(10Z³ – 10L²X+5LX² – x³) | dB
A
120L EI
30EI
24EI
M,L?
Smax
9/3EI
Mo
L
at x =
M,L
Mox
6EI
(L² – x²)
6LEI
MoL²
dcenter =
MoL
Og =
3EI
Mox?
MọL²
16EI
MoL
OB =
EI
A
Mo
2EI
2EI
L](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb768791c-c126-41b1-86e6-63f173c6340e%2F901d69b3-420f-424b-8e1c-dd9922295ee5%2F9w9m5a9_processed.png&w=3840&q=75)
Transcribed Image Text:PL³
Smax =
48EI
PL?
O, = OR =
16EI
Px
Px2
(3L — х)
6EI
PL³
dB
PL²
OB =
2EI
B
(3L² – 4x²) 0 <xs
A
48EI
х
3EI
Pb(L² – b²)³/²
9/3LEI
L² - b²
Pbx
:(L² – x² – b²)
0sxsa
Smax =
6L EI
Px2
(За — х) 0<x<a
6EI
Pab
O, =
(2L – a)
6L EI
В
Ра?
Pa?
OB
8 =
Pb [L
(x– a)³ + (L² – b²)x – x³|
6L EI b
A
at x =
8 =
SB =
(3L – a)
3
Pab
Pa?
6 ΕΙ
2EI
OR =
(2L – b)
6L EI
Pb
(3L² – 4b²) if a > b
48EI
b
(3х — а) а <х<L
6EI
a<x<L
Sœnter =
Wo
Wo
woL³
O4 = Og =
24EI
Wo.x
(L³ – 2Lx² +x³)
Smax =
Wo.x?
(6L² – 4Lx+ x²)
B
woL4
24EI
384EI
A
24EI
OB =
6ΕΙ
8EI
woX
-|a²(a – 2L)² +2a(a – 2L)x² + Lx³] | dœnter =
24L EI
384EI
Wo
(5Lª
12L²6² + 8b*)
Woa?
(a – 2L)²
84 =
wo
0<xsa
24L EI
Wox?
(6a² – 4ax+ x²) 0<x<a
if a 2 b
Woa²
Woa?
24L EI
В
(2L² – a²)
woa³
Woa?
(3L² – 2a²) if asb
96EI
24EI
woa?
(-L+x)(a² – 4Lx+ 2x²)
OR =
A
dB
(4L – a)| 0B
denter =
24LEI
24EI
6EI
a<xsL
Woa3
(4х —
24EI
а)
a <x<L
L
Smax = 0.006 522
EI
7woL3
Wox
(7L* – 10L²x² +3x*)
360L EI
360EI
wo
at x = 0.5193L
woL³
OB =
+7°M
EI
OR =
45 EI
woL4
denter = 0.006 510-
B
Wox-
(10Z³ – 10L²X+5LX² – x³) | dB
A
120L EI
30EI
24EI
M,L?
Smax
9/3EI
Mo
L
at x =
M,L
Mox
6EI
(L² – x²)
6LEI
MoL²
dcenter =
MoL
Og =
3EI
Mox?
MọL²
16EI
MoL
OB =
EI
A
Mo
2EI
2EI
L
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