For the cantilever beam with shown loading, determine the deflection at the free end. (E = 200 GPa;I = 26.9 × 106 mm*)
For the cantilever beam with shown loading, determine the deflection at the free end. (E = 200 GPa;I = 26.9 × 106 mm*)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![For the cantilever beam with shown loading, determine the deflection at the free
end. (E = 200 GPa;I = 26.9 × 106 mm*)
kN
50
y
m
A
4 m
В](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f5572ba-44dc-4763-838b-a179ea7fa1de%2F11a99669-c843-4b81-a7b8-7efdfbae7501%2Fa7kjmfr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:For the cantilever beam with shown loading, determine the deflection at the free
end. (E = 200 GPa;I = 26.9 × 106 mm*)
kN
50
y
m
A
4 m
В
![Table of Equations
Loading Configuration
Deflection
B
Px
A
y = (x - 3L)
6EI
(Px2
(x- 3a) 0SXSA
6EI
y =
Pa?
(a - 3x) a <XSL
-6EI
A
Wo
A.
y =
24ET (-612 + 4Lx – x2)
24EI
Luduing Contiguration
Deflection
Wo
(Wox?
DAE
y =
B
(-6a? + 4ax - x²) 0Sxsa
A
24EI
W,a
24E (a - 4x)
asxSL
Wo
B
120LEI-10L + 10Lx – 5Lx + x)
120LEI
y =
L.
B
Px
y =
48EI
;(4x² – 31") 0sxs;
Pbx
61. ET (-12 + x² + b2)
Pb |
lEL ET(*- a)' - (L² – b²)x + x* asxsL
B
A
y =
L.
Wo
y =
y =AR(-L' + 2Lx? – x)
24EI
y
B
U -a'(a - 21.) – 2a(a – 21.)x? – Lx'] 0 sxsa
24L EI
Wea?
241 EI (L- x)(a² – 4Lx + 2x?)
a sxSL
24L EI
Wo
y =
y= 360L EI-7L' + 10L?x? – 3x*)
360L EI](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f5572ba-44dc-4763-838b-a179ea7fa1de%2F11a99669-c843-4b81-a7b8-7efdfbae7501%2F637l3wb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Table of Equations
Loading Configuration
Deflection
B
Px
A
y = (x - 3L)
6EI
(Px2
(x- 3a) 0SXSA
6EI
y =
Pa?
(a - 3x) a <XSL
-6EI
A
Wo
A.
y =
24ET (-612 + 4Lx – x2)
24EI
Luduing Contiguration
Deflection
Wo
(Wox?
DAE
y =
B
(-6a? + 4ax - x²) 0Sxsa
A
24EI
W,a
24E (a - 4x)
asxSL
Wo
B
120LEI-10L + 10Lx – 5Lx + x)
120LEI
y =
L.
B
Px
y =
48EI
;(4x² – 31") 0sxs;
Pbx
61. ET (-12 + x² + b2)
Pb |
lEL ET(*- a)' - (L² – b²)x + x* asxsL
B
A
y =
L.
Wo
y =
y =AR(-L' + 2Lx? – x)
24EI
y
B
U -a'(a - 21.) – 2a(a – 21.)x? – Lx'] 0 sxsa
24L EI
Wea?
241 EI (L- x)(a² – 4Lx + 2x?)
a sxSL
24L EI
Wo
y =
y= 360L EI-7L' + 10L?x? – 3x*)
360L EI
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