check the following and complete the question. 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 memberar pins. El is constant.2.and BM₂A =D·1. Fixed End Moments for each member.MAD=WL²12we²125mMec McB = 0КосMoc = Mco = 0KAD = 1/Member StiffnessB= -5 x 6²12+Gm5 x 6²12K = 1/2Kas = 1/= 15 KNM4=15 KNMAssume the supports at A7AX1OBCSmI/53. Cord Rotation 4 = 4/LDue to the lateral displacement at 0 and c the card memberrotate YAO = Yce =XB4. Slope Deflection EquationsMAD = 2 E1MADMDA =Мос2탄 [20x100-Mco =2E1 (20A60.667 EI DA + 0.333 0p-0.167E1A-15= 0.8 Op +282 E1 (200 +000.667 E1 +0.338 E10A-0.167 EIA + 15Mes 2 El0-34] ++ Op-0.4 El 0+ 0.8 105. EquilibriumConditionsMOA + Moc=0• +0₁) +0.4000.667 E1 00 +- 3승) -1.467 E1 00 ++ FEMN+0200 +08 -34)..= 0.667 El OC + 0.333 0B - 0.167 ELA0.8 E100+ 0.4 51 0Mec 0.333 El to + 0.687 El 08 0.167 El A0+-15] + [0.8 E1 00 + 0.4 £₁ 0c] = 0El0.333 El OA 0.167 El A +0.333 El A-0.167 EIA +15+ 0.4 El 06 = 0Mco + 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 MAD200.667 EI OA+ 0.333 El CoMBC0.333 El + 0-667 E1 08= 0Sway EquationHA+HB + 5+6 = 0HB =HA - MAD + MOA6= Mec + Mc86PHA 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

Given:Required:Moment at joints C and D

Engineering

Civil Engineering

Determine the moment at joints C and D. Assume the supports at A and B are pins. EI is constant.)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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

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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Step 1: State the given data

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Step 2: Fixed end moment

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Step 3: Slope deflection equation

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Step 4: Equilibrium equation

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Step 5: Shear equation

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Step 6: Unknowns and internal moment

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