Using this solution, kindly expand and explain how to solve for Ray, Rby. Rcy, and Rdy getting the same answers in the solution. Thank you
Using this solution, kindly expand and explain how to solve for Ray, Rby. Rcy, and Rdy getting the same answers in the solution. Thank you
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Using this solution, kindly expand and explain how to solve for Ray, Rby. Rcy, and Rdy getting the same answers in the solution. Thank you
![8:07
1) Fined end moment
1) MEAB
- 20x(8)²
12×(8)²
⇒-1984
12
30
15
2
2) MFBA =)
20x8 12x82
2176
of
12
20
15
3) MFBC =.
-2.0x10
925
-
60x10 ➜ -735 kNm
12
8
4) MFCB => 735 kNm.
5) MFCD - JUX (6)²_ 75x (2) (4) ² - 2900 kwm.
12
10x62
6) MFDC ⇒
+ 75×(4) (2) 2
190 kwm
12
62
2) MDE ⇒ - 10x22 = -20 kNm.
Slope deflection equation
1) MAB =
MFAB + 2EI [20A +06].
1) MAB = -1984 + 4EF [OB]
→ -97103 kNm
- 11) MBA => 2176 + HEI [2013]
=) ARKAN KACA 215.55 kNm
→ TERREAK WYxV-215.55 kNm
-1ii) MBC ⇒ - 725 + GEE (200 + O₂] → TRRIEAK
6EI
~iv) MCB => 725 + GET [20c+06]
=) 18/07 from 167.05 KN
3
10
/N) MCD =
-290 + 251 [20c + OB] => -167.05 kNm
*) Moc
⇒
198 + 257 [20₂ + 0c]
20 kNm
Vii:) MDE =
20
kNm
-
||||
O
kNm
90%
kwm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62f27051-e8f0-4008-b0e4-fa1534d43f62%2Feed5c39a-df5d-4dbb-98d8-71ae42a4c8a3%2Fe97usld_processed.jpeg&w=3840&q=75)
Transcribed Image Text:8:07
1) Fined end moment
1) MEAB
- 20x(8)²
12×(8)²
⇒-1984
12
30
15
2
2) MFBA =)
20x8 12x82
2176
of
12
20
15
3) MFBC =.
-2.0x10
925
-
60x10 ➜ -735 kNm
12
8
4) MFCB => 735 kNm.
5) MFCD - JUX (6)²_ 75x (2) (4) ² - 2900 kwm.
12
10x62
6) MFDC ⇒
+ 75×(4) (2) 2
190 kwm
12
62
2) MDE ⇒ - 10x22 = -20 kNm.
Slope deflection equation
1) MAB =
MFAB + 2EI [20A +06].
1) MAB = -1984 + 4EF [OB]
→ -97103 kNm
- 11) MBA => 2176 + HEI [2013]
=) ARKAN KACA 215.55 kNm
→ TERREAK WYxV-215.55 kNm
-1ii) MBC ⇒ - 725 + GEE (200 + O₂] → TRRIEAK
6EI
~iv) MCB => 725 + GET [20c+06]
=) 18/07 from 167.05 KN
3
10
/N) MCD =
-290 + 251 [20c + OB] => -167.05 kNm
*) Moc
⇒
198 + 257 [20₂ + 0c]
20 kNm
Vii:) MDE =
20
kNm
-
||||
O
kNm
90%
kwm
![Compatability Conditions
i) EMB =0.
MBA + MBC =0
EI [2.20B + 0.60c] = 96.6
ii) {Mc =0
MCB + McD = 0
EI [0.60B + 25 0₁ +0] -145
(1)
EMD=0
MeD+ MDE = 0.
ES [ ²2/06 + 400] => -130
3
Solving
0.0, (3 що
"
get
70.479
OB
-97.424
E2
FI
Op = -16.288
ΕΣ
97-03 kNm
60KN
20 kN/m
78 KN
167.05
20 kN/m↓
1 10 kN/m
20
mmmmm) (mm)
²
D
Rm³k um x
↑
个个
125-15
30.5
=
$1.18 KN
8m
32 kN/m
215-55
1) BLE
12682 134-85
MA= 97.03 kNm
яму
= 8148 KN
||||
RBy
261.67 km.
Rey 229.65 KN
Roy
= 50.5 KN,
1645
KN.
KN
10/100/
2m
20 ka](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62f27051-e8f0-4008-b0e4-fa1534d43f62%2Feed5c39a-df5d-4dbb-98d8-71ae42a4c8a3%2Fcjdiyf7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Compatability Conditions
i) EMB =0.
MBA + MBC =0
EI [2.20B + 0.60c] = 96.6
ii) {Mc =0
MCB + McD = 0
EI [0.60B + 25 0₁ +0] -145
(1)
EMD=0
MeD+ MDE = 0.
ES [ ²2/06 + 400] => -130
3
Solving
0.0, (3 що
"
get
70.479
OB
-97.424
E2
FI
Op = -16.288
ΕΣ
97-03 kNm
60KN
20 kN/m
78 KN
167.05
20 kN/m↓
1 10 kN/m
20
mmmmm) (mm)
²
D
Rm³k um x
↑
个个
125-15
30.5
=
$1.18 KN
8m
32 kN/m
215-55
1) BLE
12682 134-85
MA= 97.03 kNm
яму
= 8148 KN
||||
RBy
261.67 km.
Rey 229.65 KN
Roy
= 50.5 KN,
1645
KN.
KN
10/100/
2m
20 ka
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