*12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6 (10³)ksi.
*12-48. The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. Ew = 1.6 (10³)ksi.
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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![Figure 12-47 Full Alternative Text
*12-48. The wooden beam is subjected to the load shown. Determine the equation of the
elastic curve. Specify the deflection at the end C. Ew 1.6 (10³)ksi.
Prob. 12-48
A
-9 ft
0.8 kip/ft
B
=
9 ft
1.5 kip
C
HE
6 in.
12 in.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2Fcde225b0-1751-4491-9be2-3b883c32bfdf%2Fcy86ii_processed.png&w=3840&q=75)
Transcribed Image Text:Figure 12-47 Full Alternative Text
*12-48. The wooden beam is subjected to the load shown. Determine the equation of the
elastic curve. Specify the deflection at the end C. Ew 1.6 (10³)ksi.
Prob. 12-48
A
-9 ft
0.8 kip/ft
B
=
9 ft
1.5 kip
C
HE
6 in.
12 in.
![3.6
6
3
9
Ay
By
EMA=0= -(6.3.6) + (9. By) -(18.1.5)
By = (6.3.6) + (18.1.5)
= 5.4 Kip
ΣFy=0= + Ay -3.6 +5.4 -1.5
Ay= 0.3 Kip
R = = (x₁) (0-0898, xJ
(0.0888 x₁)
XV
^^
0.3
X₁
EMO=O=M +
M = -0.08x₁³
0.3 X₁
El dx = -0.0888x²³ -0.3X
XI 0-0888X,
2
-0.0888 x
24
-0.0888X,5
120
0.8
0.8 = #
H
X₁
2
0.3x₁²
(X₁.0.3) Kip-ft
+ Ci
El V=SS -0.0888 X1² - 0.3 X₁² + C₁
24
2
3
0.3 X+0
6
X₁
(1
+ CIXI + C₂
H
M
1.5
(18-X2)
EMO =O=-M-((18-x₂)-1.5)
=-M-(27-1.5 X₂)
M= 1.5 X2-27
El=1.5 X₂-27
↳1.5 X2-27X2 + C3.
6
Ev= 1.5x23
6
3
El
ETV = S 15X²²-27X2² - C3X= + C4 4
1.5x₁
+
2
Elv = -0.0988X5
120
El = 15x₂² - 27 x₂ -C36
X1=0
El ay = -0.0888x₁¹_ 0.3x² + ₁
我
24
V₁=0
²-27 x2² - (3X² + C+
+C4 (4)
- 0.3 x₁ ³ + Cixi + C₂ ☺
(2) C₂=0
@ 0=-0.0888 (915-03 (95²- C₁ (1)-0
+
X₁=9|
X₁=9
Vi=O
6
80.146 = C1 (a) - Ci=8.905
dvi-dvz
dx dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2Fcde225b0-1751-4491-9be2-3b883c32bfdf%2F5thslj_processed.png&w=3840&q=75)
Transcribed Image Text:3.6
6
3
9
Ay
By
EMA=0= -(6.3.6) + (9. By) -(18.1.5)
By = (6.3.6) + (18.1.5)
= 5.4 Kip
ΣFy=0= + Ay -3.6 +5.4 -1.5
Ay= 0.3 Kip
R = = (x₁) (0-0898, xJ
(0.0888 x₁)
XV
^^
0.3
X₁
EMO=O=M +
M = -0.08x₁³
0.3 X₁
El dx = -0.0888x²³ -0.3X
XI 0-0888X,
2
-0.0888 x
24
-0.0888X,5
120
0.8
0.8 = #
H
X₁
2
0.3x₁²
(X₁.0.3) Kip-ft
+ Ci
El V=SS -0.0888 X1² - 0.3 X₁² + C₁
24
2
3
0.3 X+0
6
X₁
(1
+ CIXI + C₂
H
M
1.5
(18-X2)
EMO =O=-M-((18-x₂)-1.5)
=-M-(27-1.5 X₂)
M= 1.5 X2-27
El=1.5 X₂-27
↳1.5 X2-27X2 + C3.
6
Ev= 1.5x23
6
3
El
ETV = S 15X²²-27X2² - C3X= + C4 4
1.5x₁
+
2
Elv = -0.0988X5
120
El = 15x₂² - 27 x₂ -C36
X1=0
El ay = -0.0888x₁¹_ 0.3x² + ₁
我
24
V₁=0
²-27 x2² - (3X² + C+
+C4 (4)
- 0.3 x₁ ³ + Cixi + C₂ ☺
(2) C₂=0
@ 0=-0.0888 (915-03 (95²- C₁ (1)-0
+
X₁=9|
X₁=9
Vi=O
6
80.146 = C1 (a) - Ci=8.905
dvi-dvz
dx dx
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