Determine the moment at joints C and D. Assume the supports at A and B are pins. EI is constant.)
Determine the moment at joints C and D. Assume the supports at A and B are pins. EI is constant.)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
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Frames using Slope Deflection method
(Determine the moment at joints C and D. Assume the supports at A and B are pins. EI is constant.)
![Determined the moment in each member
ar pins. El is constant.
2.
and B
M₂A =
D
·1. Fixed End Moments for each member.
MAD=WL²
12
we²
12
5m
Mec McB = 0
Кос
Moc = Mco = 0
KAD = 1/
Member Stiffness
B
= -5 x 6²
12
+
Gm
5 x 6²
12
K = 1/2
Kas = 1/
= 15 KNM
4
=
15 KNM
Assume the supports at A
7
AX
1
OBC
Sm
I/5
3. Cord Rotation 4 = 4/L
Due to the lateral displacement at 0 and c the card member
rotate YAO = Yce =
XB
4. Slope Deflection Equations
MAD = 2 E1
MAD
MDA =
Мос
2탄 [20x100-
Mco =
2E1 (20A
6
0.667 EI DA + 0.333 0p-0.167E1A-15
= 0.8 Op +
28
2 E1 (200 +00
0.667 E1 +0.338 E10A-0.167 EIA + 15
Mes 2 El
0-34] +
+ Op-
0.4 El 0+ 0.8 10
5. Equilibrium
Conditions
MOA + Moc=0
• +0₁) +
0.400
0.667 E1 00 +
- 3승) -
1.467 E1 00 +
+ FEMN
+0
200 +08 -34)..
= 0.667 El OC + 0.333 0B - 0.167 ELA
0.8 E100+ 0.4 51 0
Mec 0.333 El to + 0.687 El 08 0.167 El A
0
+
-15] + [0.8 E1 00 + 0.4 £₁ 0c] = 0
El
0.333 El OA 0.167 El A +
0.333 El A-0.167 EIA +15+ 0.4 El 06 = 0
Mco + McB = 0
(0.4 El 00 + 0.8 E1 00) + (0.667 EI 0 + 0.333 08-0.167 El.
EIA).
0.4 51% +1.467 E10+ 0.333 0 - 0.167 EI A = 0
=O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafdb1323-93e6-46d8-8747-9f1d98e818fe%2Fc617099f-3158-457a-8a6e-f00c98f90d73%2Fo41ot6i_processed.png&w=3840&q=75)
Transcribed Image Text:Determined the moment in each member
ar pins. El is constant.
2.
and B
M₂A =
D
·1. Fixed End Moments for each member.
MAD=WL²
12
we²
12
5m
Mec McB = 0
Кос
Moc = Mco = 0
KAD = 1/
Member Stiffness
B
= -5 x 6²
12
+
Gm
5 x 6²
12
K = 1/2
Kas = 1/
= 15 KNM
4
=
15 KNM
Assume the supports at A
7
AX
1
OBC
Sm
I/5
3. Cord Rotation 4 = 4/L
Due to the lateral displacement at 0 and c the card member
rotate YAO = Yce =
XB
4. Slope Deflection Equations
MAD = 2 E1
MAD
MDA =
Мос
2탄 [20x100-
Mco =
2E1 (20A
6
0.667 EI DA + 0.333 0p-0.167E1A-15
= 0.8 Op +
28
2 E1 (200 +00
0.667 E1 +0.338 E10A-0.167 EIA + 15
Mes 2 El
0-34] +
+ Op-
0.4 El 0+ 0.8 10
5. Equilibrium
Conditions
MOA + Moc=0
• +0₁) +
0.400
0.667 E1 00 +
- 3승) -
1.467 E1 00 +
+ FEMN
+0
200 +08 -34)..
= 0.667 El OC + 0.333 0B - 0.167 ELA
0.8 E100+ 0.4 51 0
Mec 0.333 El to + 0.687 El 08 0.167 El A
0
+
-15] + [0.8 E1 00 + 0.4 £₁ 0c] = 0
El
0.333 El OA 0.167 El A +
0.333 El A-0.167 EIA +15+ 0.4 El 06 = 0
Mco + McB = 0
(0.4 El 00 + 0.8 E1 00) + (0.667 EI 0 + 0.333 08-0.167 El.
EIA).
0.4 51% +1.467 E10+ 0.333 0 - 0.167 EI A = 0
=O
![MAD
20
0.667 EI OA+ 0.333 El Co
MBC
0.333 El + 0-667 E1 08
= 0
Sway Equation
HA+HB + 5+6 = 0
HB =
HA - MAD + MOA
6
= Mec + Mc8
6
P
HA HB + 30 - 0
-
0.167 E14 - 15 = 0 Ⓒ
0.167 EID=0Ⓒ
(E10A+ El-0.333 EIA
(E1 08 + El 0c - 0.333 El A)
=> + (E10A + E108 + E1GC + E100 -0.666 El A
[Inanyats]
E¹A) + 30 =0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafdb1323-93e6-46d8-8747-9f1d98e818fe%2Fc617099f-3158-457a-8a6e-f00c98f90d73%2F7bxg9q_processed.png&w=3840&q=75)
Transcribed Image Text:MAD
20
0.667 EI OA+ 0.333 El Co
MBC
0.333 El + 0-667 E1 08
= 0
Sway Equation
HA+HB + 5+6 = 0
HB =
HA - MAD + MOA
6
= Mec + Mc8
6
P
HA HB + 30 - 0
-
0.167 E14 - 15 = 0 Ⓒ
0.167 EID=0Ⓒ
(E10A+ El-0.333 EIA
(E1 08 + El 0c - 0.333 El A)
=> + (E10A + E108 + E1GC + E100 -0.666 El A
[Inanyats]
E¹A) + 30 =0
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