For the cantilever beam with shown loading, determine the deflection at the free end. (E = 200 GPa; 1 = 26.9 x 106 mm¹) "1 A 5 m B 40 kN m x
For the cantilever beam with shown loading, determine the deflection at the free end. (E = 200 GPa; 1 = 26.9 x 106 mm¹) "1 A 5 m B 40 kN m x
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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Question
Refer to the table of equations when answering.
Transcribed Image Text:For the cantilever beam with shown loading, determine the deflection at the free
end. (E = 200 GPa;1 = 26.9 x 104 mm*)
kN
40
A
B
5 m
![Table of Equations
Loading Configuration
Deflection
B
Px
A
y =
=ZR
:(x- 3L)
6EI
(Px2
(x – 3a) 0 sIsa
6EI
Ра?
A (a - 3x) asxSL
y =
-6EI
Wo
B
A.
y =:
24EI
24ET (-612 + 4Lx - x2)
L.
Luduing Contiguration
Deflection
Wo
B
(Wox?
ZAE (-6a? + 4ax - x) 0s xsa
24EI
A
y =
W,a
24E (a - 4x)
asxSL
Wo
Wex?
120LEI-10L + 10Lx – 5Lx + x)
B
A
y =
120LEI
L.
B
Px
y =
48EI
;(4x² – 31²) 0sxs;
Pbx
61. ET -L + x² + b²)
lEL ET(*- a) - (L² – b²)x + x* asxsL
B
OSxsa
A
y =
Pb
L.
Wo
y =
y =AR(-L' + 2Lx? - x)
24EI
mim
y
B
¡|-a²(a – 21.)² – 2a(a – 21.)x² – Lx*] 0sIsa
24L EI
Wea?
241 EI (L- x)(a² – 4Lx + 2x?)
asxSL
24L EI
Wo
y = 360L EI-7L + 10L?x2 – 3x*)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f5572ba-44dc-4763-838b-a179ea7fa1de%2F13489dd4-cb85-4e51-94c8-c5f69d00c4c3%2Foy5fr3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Table of Equations
Loading Configuration
Deflection
B
Px
A
y =
=ZR
:(x- 3L)
6EI
(Px2
(x – 3a) 0 sIsa
6EI
Ра?
A (a - 3x) asxSL
y =
-6EI
Wo
B
A.
y =:
24EI
24ET (-612 + 4Lx - x2)
L.
Luduing Contiguration
Deflection
Wo
B
(Wox?
ZAE (-6a? + 4ax - x) 0s xsa
24EI
A
y =
W,a
24E (a - 4x)
asxSL
Wo
Wex?
120LEI-10L + 10Lx – 5Lx + x)
B
A
y =
120LEI
L.
B
Px
y =
48EI
;(4x² – 31²) 0sxs;
Pbx
61. ET -L + x² + b²)
lEL ET(*- a) - (L² – b²)x + x* asxsL
B
OSxsa
A
y =
Pb
L.
Wo
y =
y =AR(-L' + 2Lx? - x)
24EI
mim
y
B
¡|-a²(a – 21.)² – 2a(a – 21.)x² – Lx*] 0sIsa
24L EI
Wea?
241 EI (L- x)(a² – 4Lx + 2x?)
asxSL
24L EI
Wo
y = 360L EI-7L + 10L?x2 – 3x*)
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