Slope deflection method

Mr. S. R. Suryawanshi1
JSPM’s Imperial College of Engg. & Research,
Civil Engineering Department
Structural Analysis-II
TE Civil (2015c)
Date:-08/07/2020
Mr. S. R. Suryawanshi
Assistant Professor in Civil Engineering
JSPM’s ICOER, Wagholi, Pune
Email: srsuryawanshi_civil@jspmicoer.edu.in
Mobile : 9860079033

UNIT-I-SLOPE-DEFLECTION
METHOD OF ANALYSIS
Mr. S. R. Suryawanshi 2
A. APPLICATION TO BEAMS WITH AND
WITHOUT JOINT TRANSLATION AND
ROTATION, YIELDING OF SUPPORT.
B. NON-SWAY & SWAY ANALYSIS OF RIGID
JOINTED RECTANGULAR PORTAL
FRAMES
(INVOLVING NOT MORE THAN THREE UNKNOWNS)

Slope Deflection Equation

Steps to follow :-
1. Find Dki (degree of kinematic indeterminacy/Degree of
freedom)
2. Find fixed end moment
3. Write Slope deflection equations for each span
4. Apply the equilibrium condition at joints

5. Find the unknown displacements
6.Find final end moments
7. Find reactions and draw SFD
8. Draw Superimposed BMD
Steps to follow :-

Types of Numerical to Study:-
A. Analysis of Indeterminate Beams
1. Continuous beam with both ends fixed
2. Continuous beam with one end or both ends
simple
3. Continuous beam with overhanging span
4. Continuous beam with sinking of support

B. Analysis of Indeterminate Rigid jointed
rectangular portal frames
1. Non-sway frames
2. Sway frames.
Types of Numerical to Study:-

TE Civil (2015c)
Date:-13/07/2020
Mobile : 9860079033

Q. Analyse the beam shown in fig. using
Slope Deflection Method, Draw
superimposed BMD take EI=Constant
(10Marks)

Solution:-
Step 1] find degree of freedom (Dki)
Dki=01(𝜃 𝑏)
Step 2] Find fixed end moments
For span AB
Mab=
−𝑾×𝑳 𝟐
𝟏𝟐
=
−𝟐𝟎×𝟑 𝟐
𝟏𝟐
= -15 kN.m
Mba=
+𝑾×𝑳 𝟐
𝟏𝟐
=
+𝟐𝟎×𝟑 𝟐
𝟏𝟐
= +15 kN.m

For span BC
Mbc=
−𝑾𝑳
𝟖
=
−𝟓𝟎×𝟒
𝟖
= -25kN.m
Mcb=
+𝑾𝑳
𝟖
=
+𝟓𝟎×𝟒
𝟖
= +25kN.m

Step 3] write slope deflection equations
For span AB
𝑀 𝑎𝑏 = 𝑀 𝑎𝑏 +
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
)
as No support settlement so,
3∆
𝐿
= 𝟎
2𝐸𝐼
𝐿
2𝜃 𝑎 + 𝜃 𝑏
= −15 +
2𝐸𝐼
3
( 2𝜃 𝑎 + 𝜃 𝑏)

As Support A is fixed 𝜃 𝑎 = 0
𝐌 𝐚𝐛 = −𝟏𝟓 +
𝟐𝐄𝐈
𝟑
𝛉 𝐛
𝐌 𝐚𝐛 = −𝟏𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 ……… (1)

𝑴 𝒃𝒂 = 𝑴 𝒃𝒂 +
𝟐𝑬𝑰
𝑳
𝜽 𝒂 + 𝟐𝜽 𝒃
=𝟏𝟓 +
𝟐𝑬𝑰
𝟑
𝐌 𝐛𝐚 = 𝟏𝟓 +
𝟐𝐄𝐈
𝟑
𝟐𝛉 𝐛
𝐌 𝐛𝐚= 𝟏𝟓 + 𝟏. 𝟑𝟑𝐄𝐈𝛉 𝐛 ………… (2)

Now considering span BC
𝑴 𝒃𝒄 = 𝑴 𝒃𝒄 +
𝟐𝑬𝑰
𝑳
𝟐𝜽 𝒃 + 𝜽 𝒄
=−𝟐𝟓 +
𝟐𝑬𝑰
𝟒
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
As Support C is fixed 𝜃𝑐 = 0
𝐌 𝐛𝐜 = −𝟐𝟓 +
𝟐𝐄𝐈
𝟒
𝟐𝛉 𝐛
𝐌 𝐛𝐜 = −𝟐𝟓 + 𝐄𝐈𝛉 𝐛 ……… (3)

𝑴 𝒄𝒃 = 𝑴 𝒄𝒃 +
𝟐𝑬𝑰
𝑳
𝜽 𝒃 + 𝟐𝜽 𝒄
= 𝟐𝟓 +
𝟐𝑬𝑰
𝟒
( 𝜽 𝒃 + 𝟐𝜽 𝒄)
𝑴 𝒄𝒃 = 𝟐𝟓 +
𝟐𝑬𝑰
𝟒
𝜽 𝒃
𝑴 𝒄𝒃 = 𝟐𝟓 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 …………….(4)

Step 4] Apply joint equilibrium condition at
Joint B
At Joint B
i.e. Equation No (2) + Equation No. (3) = 0
𝟏𝟓 + 𝟏. 𝟑𝟑𝐄𝐈𝛉 𝐛 + −𝟐𝟓 + 𝐄𝐈𝛉 𝐛 = 𝟎

By solving above we get
𝟏. 𝟑𝟑 + 𝟏 𝐄𝐈𝛉 𝐛 = 𝟐𝟓 − 𝟏𝟓
𝟐. 𝟑𝟑𝐄𝐈𝛉 𝐛 = 𝟏𝟎
..unknown displacement

Step 5] Find Final End moments
Put the value of EIθb = 4.291in Equation
(1), (2), (3) & (4) to get final
End Moments
𝐌 𝐚𝐛 = −𝟏𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛
= −15 + 0.67 × 4.291
= −12.125𝑘𝑁. 𝑚

𝐌 𝐛𝐚 = 15 + 1.33𝐄𝐈𝛉 𝐛
= 15 + 1.33 × 4.291 = 20.709kN. m
𝐌 𝐛𝐜 = −𝟐𝟓 + 𝐄𝐈𝛉 𝐛
= −25 + 4.291 = −20.709kN. m
𝑴 𝒄𝒃 = 𝟐𝟓 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃
= 25 + 0.5 × 4.291 = 27.145kN. m

(In kN.m)21
Step 6] Superimposed BMD
3m 2m 2m
WL2/8=20 x 32 /8
=22.50
WL/4=50x4/4
=50
12.125
20.709
27.145
A B C

TE Civil (2015c)
Date:-14/07/2020
Mobile : 9860079033

Quick Revision of Last Session
Discussed Type-I numerical i.e.
Continuous beam with both
ends fixed

Unit-I:- Slope Deflection Method of
Analysis
A] Analysis of statically Indeterminate
Beams
Type II] Continuous beam with end
support simple

SPPU Insem Exam
Aug.2017

Solution:-
Dki=02(𝜃 𝑏,𝜃𝑐)
For span AB
Mab=
−𝑾×𝑳 𝟐
𝟏𝟐
=
−𝟐𝟎×𝟑 𝟐
𝟏𝟐
= -15 kN.m
Mba=
+𝑾×𝑳 𝟐
𝟏𝟐
=
+𝟐𝟎×𝟑 𝟐
𝟏𝟐
= +15 kN.m

For span BC
Mbc=
−𝑾𝑳
𝟖
=
−𝟓𝟎×𝟒
𝟖
= -25kN.m
Mcb=
+𝑾𝑳
𝟖
=
+𝟓𝟎×𝟒
𝟖
= +25kN.m

For span AB
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
)
as No support settlement so,
3∆
𝐿
= 𝟎
2𝐸𝐼
𝐿
= −15 +
2𝐸𝐼
3
( 2𝜃 𝑎 + 𝜃 𝑏)

𝐌 𝐚𝐛 = −𝟏𝟓 +
𝟐𝐄𝐈
𝟑
𝛉 𝐛
𝐌 𝐚𝐛 = −𝟏𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 ……… (1)

𝟐𝑬𝑰
𝑳
=𝟏𝟓 +
𝟐𝑬𝑰
𝟑
𝐌 𝐛𝐚 = 𝟏𝟓 +
𝟐𝐄𝐈
𝟑
𝟐𝛉 𝐛
𝐌 𝐛𝐚= 𝟏𝟓 + 𝟏. 𝟑𝟑𝐄𝐈𝛉 𝐛 ………… (2)

𝟐𝑬𝑰
𝑳
=−𝟐𝟓 +
𝟐𝑬𝑰
𝟒
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
As Support C is simple 𝜃𝑐 ≠ 0
𝐌 𝐛𝐜 = −𝟐𝟓 +
𝟐𝐄𝐈
𝟒
𝟐𝛉 𝐛 + 𝛉 𝒄
𝐌 𝐛𝐜 = −𝟐𝟓 + 𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄……… (3)

𝟐𝑬𝑰
𝑳
= 𝟐𝟓 +
𝟐𝑬𝑰
𝟒
( 𝜽 𝒃 + 𝟐𝜽 𝒄)
𝑴 𝒄𝒃 = 𝟐𝟓 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 …….(4)

Joint B
At Joint B
𝟏𝟓 + 𝟏. 𝟑𝟑𝐄𝐈𝛉 𝐛 + −𝟐𝟓 + 𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄
= 𝟎

𝟏. 𝟑𝟑 + 𝟏 𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄 = 𝟐𝟓 − 𝟏𝟓
𝟐. 𝟑𝟑𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄 = 𝟏𝟎 … … . . (𝟓)
At Joint C
i.e. Equation No.(4)=0
𝟐𝟓 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄=0
𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 = -25 ……….(6)

By solving Equation No.(5) and (6) we get
𝟐. 𝟑𝟑𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄 = 𝟏𝟎 … … . . 𝟓
𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 = -25 ……….(6)

Put the value of EIθb = 10.817 & EIθc =
− 30.408 in Equation (1), (2), (3) & (4) to
get final End Moments,
𝐌 𝐚𝐛 = −𝟏𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛
= −15 + 0.67 × 10.817
= −7.752𝑘𝑁. 𝑚

38
𝑴 𝐛𝐚 = 𝟏𝟓 + 𝟏. 𝟑𝟑𝐄𝐈𝛉 𝐛
= 𝟏𝟓 + 𝟏. 𝟑𝟑 × 𝟏𝟎. 𝟖𝟏𝟕 = 𝟐𝟗. 𝟑𝟖𝟕𝐤𝐍. 𝐦
𝑴 𝒃𝒄 = −𝟐𝟓 + 𝑬𝑰𝜽 𝒃 + 𝟎. 𝟓𝑬𝑰𝜽 𝒄
= −𝟐𝟓 + 𝟏𝟎. 𝟖𝟏𝟕 + 𝟎. 𝟓 × −𝟑𝟎. 𝟒𝟎𝟖
= −𝟐𝟗. 𝟑𝟖𝟕𝐤𝐍. 𝐦
𝑴 𝒄𝒃 = 𝟐𝟓 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄
= 𝟐𝟓 + 𝟎. 𝟓 × 𝟏𝟎. 𝟖𝟏𝟕 + −𝟑𝟎. 𝟒𝟎𝟖
= 𝟎

(In kN.m)39
3m 2m 2m
WL2/8=20 x 32 /8
=22.50
WL/4=50x4/4
=50
7.752
29.387
A B C

TE Civil (2015c)
Date:-15/07/2020
Mobile : 9860079033

Discussed Numerical on
Continuous beam with End
simple

Analysis
Beams
MCQs from Competitive examination &
Type II] Continuous beam with end
support simple

47

Slope Deflection Method, Draw superimposed
BMD take EI=Constant (10Marks)

49
Solution:-
Dki=03(𝜃 𝑏,𝜃𝑐 , 𝜃 𝑑)
For span AB
Mab=
−𝑾𝑳
𝟖
=
−𝟕𝟓×𝟔
𝟖
= -56.25kN.m
Mba=
+𝑾𝑳
𝟖
=
+𝟕𝟓×𝟔
𝟖
= +56.25kN.m

50
For span BC
Mbc=
−𝑾×𝑳 𝟐
𝟏𝟐
=
−𝟖𝟓×𝟔 𝟐
𝟏𝟐
= -255 kN.m
Mcb=
+𝑾×𝑳 𝟐
𝟏𝟐
=
+𝟖𝟓×𝟔 𝟐
𝟏𝟐
= +255 kN.m
For span CD
Mcd=
−𝑾×𝒂×𝒃 𝟐
𝑳 𝟐 =
−𝟓𝟎×𝟐×𝟒 𝟐
𝟔 𝟐 = -44.44 kN.m
MdC=
+𝑾×𝒂 𝟐
×𝒃
𝑳 𝟐 =
+𝟓𝟎×𝟐 𝟐
×𝟒
𝟔 𝟐 == +22.22 kN.m

For span AB
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏)
2𝐸𝐼
𝐿
= −15 +
2𝐸𝐼
3
( 2𝜃 𝑎 + 𝜃 𝑏)

𝐌 𝐚𝐛 = −𝟓𝟔. 𝟐𝟓 +
𝟐𝐄𝐈
𝟔
𝛉 𝐛
𝐌 𝐚𝐛 = −𝟓𝟔. 𝟐𝟓 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝐛 ……… (1)

𝟐𝑬𝑰
𝑳
=𝟓𝟔. 𝟐𝟓 +
𝟐𝑬𝑰
𝟔
𝐌 𝐛𝐚 = 𝟓𝟔. 𝟐𝟓 +
𝟐𝐄𝐈
𝟔
𝟐𝛉 𝐛
𝐌 𝐛𝐚= 𝟓𝟔. 𝟐𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 …… (2)

𝟐𝑬𝑰
𝑳
=−𝟐𝟓𝟓 +
𝟐𝑬𝑰
𝟔
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
𝐌 𝐛𝐜 = −𝟐𝟓𝟓 +
𝟐𝐄𝐈
𝟔
𝐌 𝐛𝐜 = −𝟐𝟓𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒄…(3)

𝟐𝑬𝑰
𝑳
= 𝟐𝟓𝟓 +
𝟐𝑬𝑰
𝟔
( 𝜽 𝒃 + 𝟐𝜽 𝒄)
𝑴 𝒄𝒃 = 𝟐𝟓𝟓 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒄 ...(4)

Now considering span CD
𝑴 𝒄𝒅 = 𝑴 𝒄𝒅 +
𝟐𝑬𝑰
𝑳
𝟐𝜽 𝒄 + 𝜽 𝒅
=−𝟒𝟒. 𝟒𝟒 +
𝟐𝑬𝑰
𝟔
( 𝟐𝜽 𝒄 + 𝜽 𝒅)
𝐌 𝒄𝒅 = −𝟒𝟒. 𝟒𝟒 +
𝟐𝐄𝐈
𝟔
𝟐𝛉 𝒄 + 𝛉 𝒅
𝐌 𝒄𝒅 = −𝟒𝟒. 𝟒𝟒 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝒄 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒅…(5)

𝑴 𝒅𝒄 = 𝑴 𝒅𝒄 +
𝟐𝑬𝑰
𝑳
𝜽 𝒄 + 𝟐𝜽 𝒅
= 𝟐𝟐. 𝟐𝟐 +
𝟐𝑬𝑰
𝟔
( 𝜽 𝒄 + 𝟐𝜽 𝒅)
𝑴 𝒅𝒄 = 𝟐𝟐. 𝟐𝟐 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒅 ...(6)

Joint B
At Joint B
𝟓𝟔. 𝟐𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 + −𝟐𝟓𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒄
= 𝟎

59
𝟏. 𝟑𝟒𝐄𝐈𝛉 𝐛 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒄 = 𝟏𝟗𝟖. 𝟕𝟒 … … . . (𝟕)
At Joint C
i.e. Equation No.(4) + Equation No.(5) =0
𝟐𝟓𝟓 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒄 +[ −𝟒𝟒. 𝟒𝟒 +
𝟎. 𝟔𝟕𝐄𝐈𝛉 𝒄 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒅]=0
𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃 + 𝟏. 𝟑𝟒𝑬𝑰𝜽 𝒄+𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒅 = -210.56
……….(8)

At Joint D
i.e. Equation No.(6) =0
𝟐𝟐. 𝟐𝟐 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒅=0
𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒅= -22.22 ……….(9)

By solving Equation No.(7) , (8) & (9) we get
𝟏. 𝟑𝟒𝑬𝑰𝜽 𝒃 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 = 𝟏𝟗𝟖. 𝟕𝟒
𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃 + 𝟏. 𝟑𝟒𝑬𝑰𝜽 𝒄+𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒅 = -210.56
𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒅= -22.22

Put the value of EIθb = 204.154 , EIθc =
− 226.747 &EIθ 𝑑 = 78.517 in Equation (1),
(2), (3) , (4),(5)& (6) to get final End
Moments,

63
𝐌 𝐚𝐛 = −𝟓𝟔. 𝟐𝟓 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝐛
= 11.120 𝑘𝑁. 𝑚
𝑴 𝐛𝐚 = 𝟓𝟔. 𝟐𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛
= 𝟏𝟗𝟑. 𝟎𝟑𝟓𝐤𝐍. 𝐦
𝑴 𝒃𝒄 = −𝟐𝟓𝟓 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒄
= −𝟏𝟗𝟑. 𝟎. 𝟑𝟓𝐤𝐍. 𝐦
𝑴 𝒄𝒃 = 𝟐𝟓𝟓 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒄
= 𝟏𝟕𝟎. 𝟒𝟓𝟏
𝐌 𝒄𝒅 = −𝟒𝟒. 𝟒𝟒 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝒄 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝒅
= -170.451
𝑴 𝒅𝒄 = 𝟐𝟐. 𝟐𝟐 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒄 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒅=0

TE Civil (2015c)
Date:-20/07/2020
Mobile : 9860079033

Discussed Numerical on
Continuous beam with End
simple

Analysis
Beams
Type III] Continuous beam with
overhanging span

At Simple end support C Mcb +Mcd=0
Mcd=Moment due to overhanging span
=40×2=80kN.m

Slope Deflection Method, Draw superimposed
BMD take EI=Constant (10Marks)

69
Solution:-
Dki=02(𝜃 𝑏,𝜃𝑐 )
For span AB
Mab=
−𝑾×𝑳 𝟐
𝟏𝟐
=
−𝟐𝟎×𝟔 𝟐
𝟏𝟐
= -60 kN.m
Mba=
+𝑾×𝑳 𝟐
𝟏𝟐
=
+𝟐𝟎×𝟔 𝟐
𝟏𝟐
= +60 kN.m

70
For span BC
Mbc=
−𝑾𝑳
𝟖
=
−𝟔𝟎×𝟒
𝟖
= -30 kN.m
Mcb=
+𝑾𝑳
𝟖
=
+𝟔𝟎×𝟒
𝟖
= +30 kN.m
For span CD
Mcd =Moment due to overhanging span
=40×2=80kN.m

For span AB
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏)
2𝐸𝐼
𝐿
= −60 +
2𝐸𝐼
6
( 2𝜃 𝑎 + 𝜃 𝑏)

𝐌 𝐚𝐛 = −𝟔𝟎 +
𝟐𝐄𝐈
𝟔
𝛉 𝐛
𝐌 𝐚𝐛 = −𝟔𝟎 + 𝟎. 𝟑𝟑𝐄𝐈𝛉 𝐛 ……… (1)

𝟐𝑬𝑰
𝑳
=𝟔𝟎 +
𝟐𝑬𝑰
𝟔
𝐌 𝐛𝐚 = 𝟔𝟎 +
𝟐𝐄𝐈
𝟔
𝟐𝛉 𝐛
𝐌 𝐛𝐚= 𝟔𝟎 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 …… (2)

𝟐𝑬𝑰
𝑳
=−𝟑𝟎 +
𝟐𝑬𝑰
𝟒
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
𝐌 𝐛𝐜 = −𝟑𝟎 +
𝟐𝐄𝐈
𝟒
𝐌 𝐛𝐜 = −𝟑𝟎 + 𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄…(3)

𝟐𝑬𝑰
𝑳
= 𝟑𝟎 +
𝟐𝑬𝑰
𝟒
( 𝜽 𝒃 + 𝟐𝜽 𝒄)
𝑴 𝒄𝒃 = 𝟑𝟎 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 ...(4)

Joint B
At Joint B
𝟔𝟎 + 𝟎. 𝟔𝟕𝐄𝐈𝛉 𝐛 + −𝟑𝟎 + 𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄
= 𝟎

77
𝟏. 𝟔𝟕𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄 = −𝟑𝟎 … … . . . . (𝟓)
At Joint C
i.e. Equation No.(4) + Mcd=0
𝟑𝟎 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄+[𝟖𝟎]=0
𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 = -110……..…….(6)

By solving Equation No.(5) &(6) ,we get
𝟏. 𝟔𝟕𝐄𝐈𝛉 𝐛 + 𝟎. 𝟓𝐄𝐈𝛉 𝒄 = −𝟑𝟎
𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄 = -110

Put the value of EIθb = 17.605 ,
EIθc = −118.802 in Equation (1), (2), (3)
&(4) to get final End Moments,

80
𝑴 𝒂𝒃 = −𝟔𝟎 + 𝟎. 𝟑𝟑𝑬𝑰𝜽 𝒃
= −𝟓𝟒. 𝟏𝟗𝟎 𝒌𝑵. 𝒎
𝑴 𝐛𝐚 = 𝟔𝟎 + 𝟎. 𝟔𝟕𝑬𝑰𝜽 𝒃
= 𝟕𝟏. 𝟕𝟗𝟓 𝐤𝐍. 𝐦
𝑴 𝒃𝒄 = −𝟑𝟎 + 𝑬𝑰𝜽 𝒃 + 𝟎. 𝟓𝑬𝑰𝜽 𝒄
= −𝟕𝟏. 𝟕𝟗𝟓 𝒌𝑵. 𝒎
𝑴 𝒄𝒃 = 𝟑𝟎 + 𝟎. 𝟓𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄
= −80 kN.m
𝑴 𝒄𝒅 =40×2
= 80kN.m

(In kN.m)81
6m 2m 2m
WL2/8=20 x 62 /8
=90
WL/4=60x4/4
=60
54.190
71.795
A B C D2m
80

TE Civil (2015c)
Date:-21/07/2020
Mobile : 9860079033

Analysis
Beams
Type IV] Continuous beam with
sinking of Support / Settlement of
support

88
Solution:-
Dki=02(𝜃 𝑏,𝜃𝑐 )
For span AB
Mab
−𝑊×𝑎×𝑏2
𝐿2 =
−60×3×22
52 =-28.8kN.m
Mba
−𝑊×𝑏×𝑎2
𝐿2 =
−60×2×32
52 =+43.20kN.m

89
For span BC
Mbc=
−𝑾𝑳
𝟖
=
−𝟖𝟎×𝟒
𝟖
= -40 kN.m
Mcb=
+𝑾𝑳
𝟖
=
+𝟖𝟎×𝟒
𝟖
= +40 kN.m

For span AB
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
)
2𝐸𝐼
𝐿
2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
= −28.80 +
2𝐸𝐼
5
( 2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
)

𝐌 𝐚𝐛 = −28.80 +
2𝐸𝐼
5
(𝜃 𝑏 −
3∆
𝐿
)
= −28.80 +
2 × 20000
5
𝜃 𝑏 −
3 0.005
5
=-28.80+8000𝜃 𝑏-24
𝐌 𝐚𝐛 = −𝟓𝟐. 𝟖𝟎 + 𝟖𝟎𝟎𝟎𝛉 𝐛 ……… (1)

𝑀 𝑏𝑎 = 𝑀 𝑏𝑎 +
2𝐸𝐼
𝐿
𝜃 𝑎 + 2𝜃 𝑏 −
3∆
𝐿
=43.2 +
2×20000
5
2𝜃 𝑏 −
3 0.005
5
𝐌 𝐛𝐚 = 𝟒𝟑. 𝟐 +16000𝛉 𝐛-24
𝑴 𝒃𝒂= 𝟏𝟗. 𝟐 + 𝟏𝟔𝟎𝟎𝟎𝜽 𝒃 …… (2)

𝟐𝑬𝑰
𝑳
𝟐𝜽 𝒃 + 𝜽 𝒄 −
3∆
𝐿
=−𝟒𝟎 +
𝟐×𝟐𝟎𝟎𝟎𝟎
𝟒
𝟐𝜽 𝒃 + 𝜽 𝒄 −
3 −0.005
4
𝐌 𝐛𝐜 = −𝟒𝟎 +20000𝛉 𝐛 +10000𝛉 𝐜 +37.5
𝐌 𝐛𝐜 = −𝟐. 𝟓 + 𝟐𝟎𝟎𝟎𝟎𝛉 𝐛 + 𝟏𝟎𝟎𝟎𝟎𝛉 𝒄…(3)

𝟐𝑬𝑰
𝑳
𝜽 𝒃 + 𝟐𝜽 𝒄 −
3∆
𝐿
= 𝟒𝟎 +
𝟐 × 𝟐𝟎𝟎𝟎𝟎
𝟒
( 𝜽 𝒃 + 𝟐𝜽 𝒄 −
3 −0.005
4
)
=𝟒𝟎 + 𝟏𝟎000𝛉 𝐛 + 𝟐0000𝛉 𝐜 +37.5
𝑴 𝒄𝒃 = 𝟕𝟕. 𝟓 + 𝟏𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒄 ...(4)

Joint B
At Joint B
𝟏𝟗. 𝟐 + 𝟏𝟔𝟎𝟎𝟎𝜽 𝒃 + −𝟐. 𝟓 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟏𝟎𝟎𝟎𝟎𝜽 𝒄
= 𝟎

96
𝟑𝟔𝟎𝟎𝟎𝛉 𝐛 + 𝟏𝟎𝟎𝟎𝟎𝛉 𝒄 = −𝟏𝟔. 𝟕 … … . . . . (𝟓)
At Joint C
i.e. Equation No.(4) =0 as Support C is simple
end support
𝟕𝟕. 𝟓 + 𝟏𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒄=0
𝟏𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒄= -77.5……..…….(6)

By solving Equation No.(5) &(6) ,we get
𝟑𝟔𝟎𝟎𝟎𝛉 𝐛 + 𝟏𝟎𝟎𝟎𝟎𝛉 𝒄 = −𝟏𝟔. 𝟕
𝟏𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒄= -77.5

Put the value of θb = 𝟕. 𝟏𝟏 × 𝟏𝟎−𝟒,
θc = −𝟒. 𝟐𝟑 × 𝟏𝟎−𝟑
in Equation (1), (2),
(3) &(4) to get final End Moments,

99
𝑴 𝒂𝒃 = −𝟓𝟐. 𝟖𝟎 + 𝟖𝟎𝟎𝟎𝛉 𝐛
= −𝟒𝟕. 𝟏𝟏𝟐𝒌𝑵. 𝒎
𝑴 𝐛𝐚 = 𝟏𝟗. 𝟐 + 𝟏𝟔𝟎𝟎𝟎𝜽 𝒃
= 𝟑𝟎. 𝟓𝟕𝟔𝐤𝐍. 𝐦
𝑴 𝒃𝒄 = −𝟐. 𝟓 + 𝟐𝟎𝟎𝟎𝟎𝛉 𝐛 + 𝟏𝟎𝟎𝟎𝟎𝛉 𝒄
= −𝟑𝟎. 𝟓𝟕𝟔𝒌𝑵. 𝒎
𝑴 𝒄𝒃 = 𝟕𝟕. 𝟓 + 𝟏𝟎𝟎𝟎𝟎𝜽 𝒃 + 𝟐𝟎𝟎𝟎𝟎𝜽 𝒄
=0

TE Civil (2015c)
Date:-22/07/2020
Mobile : 9860079033

Quick Revision on Analysis of
beam

I
II

III
IV

Analysis
Rigid jointed frame
Sway and Non sway Frame

What is Sway and Non-sway Frame?
Sway-
Sway Frame- displacement due to
horizontal force
move or cause to move
slowly or backwards and
forwards or from side to
side.

Sway and Non sway frame
Mr. S. R. Suryawanshi 106

Frame said to be Non-sway if :-
1.Loading on frame is symmetrical.
2. Support conditions of frames are
same.
3. Coumns are idnetical i.e. Same
E,I,A.

108
Frame said to be Non-sway if :-
*Constraints(support) at beam level
Non-Sway Frames

109
Non-Sway Frames

111
Sway Frames Sway Frames

TE Civil (2015c)
Date:-07/08/2020
Mobile : 9860079033

Analysis
B] Analysis of statically Indeterminate
Rigid jointed frame
Sway and Non sway Frame

Non-Sway Frame114
Q. Analyse the
frame shown in
Fig. Using Slope
deflection
Method Take
EI=Const.

Step 1] find degree of
freedom (Dki)
Dki=02(𝜃 𝑏,𝜃𝑐)
Step 2] Find fixed end
moments
For span AB
Mab=
−𝑾×𝑳 𝟐
𝟏𝟐
=
−𝟐𝟎×𝟒 𝟐
𝟏𝟐
= -26.67 kN.m
Mba=
+𝑾×𝑳 𝟐
𝟏𝟐
=
+𝟐𝟎×𝟒 𝟐
𝟏𝟐
= +26.67 kN.m

Mbc =
−𝐖×𝐚×𝒃 𝟐
𝐋 𝟐 =
−𝟒𝟎×𝟐×𝟑 𝟐
𝟓 𝟐
=-28.8kN.m
Mcb=
+𝐖×𝐛×𝒂 𝟐
𝐋 𝟐 =
+𝟒𝟎×𝟑×𝟐 𝟐
𝟓 𝟐
=+19.20kN.m
Mbd=
−𝑾𝑳
𝟖
=
−𝟐𝟎×𝟒
𝟖
= -10 kN.m
Mdb=
+𝑾𝑳
𝟖
=
+𝟐𝟎×𝟒
𝟖
= +10 kN.m

For span AB
𝑴 𝒂𝒃 = 𝑴 𝒂𝒃 +
𝟐𝑬𝑰
𝑳
( 𝟐𝜽 𝒂 + 𝜽 𝒃)
As support A is fixed so 𝜽 𝒂=0
= −𝟐𝟔. 𝟔𝟕 +
𝟐𝑬𝑰
𝟒
(𝜽 𝒃)
=-26.67+0.5𝑬𝑰𝜽 𝒃………..(1)

𝟐𝑬𝑰
𝑳
( 𝟐𝜽 𝒂 + 𝟐𝜽 𝒃)
=𝑴 𝒃𝒂 +
𝟐𝑬𝑰
𝟒
(𝟐𝜽 𝒃)
=+26.67+𝐄𝐈𝛉 𝐛………………….(2)
𝟐𝑬𝑰
𝑳
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
=𝑴 𝒃𝒄 +
𝟐𝑬𝑰
𝟓
( 𝟐𝜽 𝒃 + 𝜽 𝒄)
=-28.20+0.8𝐄𝐈𝛉 𝐛+0.4𝐄𝐈𝛉 𝐜…………(3)

119
𝟐𝑬𝑰
𝑳
(𝜽 𝒃 + 𝟐𝜽 𝒄)
=𝑴 𝒄𝒃 +
𝟐𝑬𝑰
𝟓
( 𝜽 𝒃 + 𝟐𝜽 𝒄)
= +19.20 + 0.𝟒𝐄𝐈𝛉 𝐛 + 0.𝟖𝐄𝐈𝛉 𝐜…………(4)
𝑴 𝒃𝒅 = 𝑴 𝒃𝒅 +
𝟐𝑬𝑰
𝑳
( 𝟐𝜽 𝒃 + 𝜽 𝒅)
=𝑴 𝒃𝒅 +
𝟐𝑬𝑰
𝟒
( 𝟐𝜽 𝒃 + 𝜽 𝒅) , 𝜽 𝒅=0 due to fixity
=-10+𝐄𝐈𝛉 𝐛 …………(5)

𝑴 𝒅𝒃 = 𝑴 𝒅𝒃 +
𝟐𝑬𝑰
𝑳
( 𝜽 𝒃 + 𝟐𝜽 𝒅)
=𝑴 𝒅𝒃 +
𝟐𝑬𝑰
𝟒
( 𝜽 𝒃 + 𝟐𝜽 𝒅) , 𝜽 𝒅=0 due to fixity
=10+𝟎. 𝟓𝐄𝐈𝛉 𝐛 …………(6)

Joint B & Joint C
Joint B
Mba+Mbc+Mbd =0
Joint C Mcb=0

Joint B Mba+Mbc+Mbd =0
i.e. Eqn (2)+ Eqn (3)+ Eqn (5)=0
26.67+ EIθb +[-28.20+0.8 𝐸𝐼𝜃 𝑏 +0.4 𝐸𝐼𝜃𝑐] +
[−10+EIθb] = 0
2.8𝐄𝐈𝛉 𝐛+0.4𝐄𝐈𝛉 𝐜=11.53……………(7)
Joint C Mcb=0 i.e. Eqn (4)=0
19.20 + 0.𝟒𝐄𝐈𝛉 𝐛 + 0.𝟖𝐄𝐈𝛉 𝐜=0
0.𝟒𝐄𝐈𝛉 𝐛 + 0.𝟖𝐄𝐈𝛉 𝐜=-19.20………..(8)

2.8𝐄𝐈𝛉 𝐛+0.4𝐄𝐈𝛉 𝐜=11.53……………(7)
0.𝟒𝐄𝐈𝛉 𝐛 + 0.𝟖𝐄𝐈𝛉 𝐜=-19.20………..(8)
By solving Eqn (7) & (8) We get,
𝐄𝐈𝛉 𝐛=8.126
𝐄𝐈𝛉 𝐜=-28.063

Put the value of EIθb = 8.126 & EIθc =
− 28.063 in Equation (1), (2), (3) , (4),(5)
&(6) to get final End Moments,

Mab=-26.67+0.5𝑬𝑰𝜽 𝒃=-22.607kN.m
Mba==+26.67+𝑬𝑰𝜽 𝒃=34.796kN.m
Mbc=-28.20+0.8𝑬𝑰𝜽 𝒃+0.4𝑬𝑰𝜽 𝒄=-32.924kN.m
Mcb=+19.20 + 0.𝟒𝑬𝑰𝜽 𝒃 + 0.𝟖𝑬𝑰𝜽 𝒄=0
Mbd==-10+𝑬𝑰𝜽 𝒃 =-1.874kN.m
Mdb=10+𝟎. 𝟓𝑬𝑰𝜽 𝒃=14.063kN.m

Step 5] Draw Superimposed BMD𝟒𝟎 × 𝟐 × 𝟑
𝟓
= 𝟒𝟖
𝟐𝟎×𝟒 𝟐
𝟖
=40
𝟐𝟎 × 𝟒
𝟒
= 𝟐𝟎
34.796
22.607
32.924
1.874
14.063
A B C
D
4m
4m
3m2m

TE Civil (2015c)
Date:-07/08/2020
Mobile : 9860079033

Sway Frame130
Q. Analyse the
frame shown in
Fig. Using Slope
deflection
Method Take
EI=Const.

For span AB
2𝐸𝐼
𝐿
( 2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
)
2𝐸𝐼
𝐿
2𝜃 𝑎 + 𝜃 𝑏 −
3∆
𝐿
=0 +
2𝐸𝐼
2
( 0 + 𝜃 𝑏 −
3∆
2
)
=𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆……………….(1)

𝟐𝑬𝑰
𝑳
( 𝜽 𝒂 + 𝟐𝜽 𝒃 −
𝟑∆
𝑳
)
𝟐𝑬𝑰
𝟐
𝟎 + 𝟐𝜽 𝒃 −
𝟑∆
𝑳
=𝟎 +
𝟐𝑬𝑰
𝟐
( 𝟎 + 𝟐𝜽 𝒃 −
𝟑∆
𝟐
)
=𝟐𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆……………….(2)

For span BC
𝑀 𝑏𝑐 = 𝑀 𝑏𝑐 +
2𝐸𝐼
𝐿
( 2𝜃 𝑏 + 𝜃𝑐 −
3∆
𝐿
)
𝑀 𝑏𝑐 = 0 +
2𝐸𝐼
2
2𝜃 𝑏 + 𝜃𝑐 −
3∆
𝐿
=0 +
2𝐸𝐼
2
(2𝜃 𝑏 + 𝜃𝑐)
=𝟐𝑬𝑰𝜽 𝒃 + 𝑬𝑰𝜽 𝒄……………….(3)
Don’t consider sway
effect in Beam

𝑀𝑐𝑏 = 𝑀𝑐𝑏 +
2𝐸𝐼
𝐿
( 𝜃 𝑏 + 2𝜃𝑐 −
3∆
𝐿
)
𝑀𝑐𝑏 = 0 +
2𝐸𝐼
2
𝜃 𝑏 + 2𝜃𝑐 −
3∆
𝐿
=0 +
2𝐸𝐼
2
(𝜃 𝑏 + 2𝜃𝑐)
=𝑬𝑰𝜽 𝒃 + 𝟐𝑬𝑰𝜽 𝒄……………….(4)

For span CD
𝑀𝑐𝑑 = 𝑀𝑐𝑑 +
2𝐸𝐼
𝐿
( 2𝜃𝑐 + 𝜃 𝑑 −
3∆
𝐿
)
𝑀𝑐𝑑 = 𝑀𝑐𝑑 +
2𝐸𝐼
1
2𝜃𝑐 + 𝜃 𝑑 −
3∆
𝐿
=0 +
2𝐸𝐼
1
(2𝜃𝑐 + 0 −
3∆
1
)
=4𝐸𝐼𝜽 𝒄 − 𝟔𝑬𝑰∆……………….(5)

𝑀 𝑑𝑐 = 𝑀 𝑑𝑐 +
2𝐸𝐼
𝐿
( 𝜃𝑐 + 2𝜃 𝑑 −
3∆
𝐿
)
𝑀 𝑑𝑐 = 𝑀 𝑑𝑐 +
2𝐸𝐼
1
𝜃𝑐 + 2𝜃 𝑑 −
3∆
𝐿
=0 +
2𝐸𝐼
1
(𝜃𝑐 + 0 −
3∆
1
)
=𝟐𝑬𝑰𝜽 𝒄 − 𝟔𝑬𝑰∆……………….(6)

Joint B , Joint C & Horizontal shear Eqllm.
At Joint B Mba+Mbc=0
At joint C Mcb+ Mcd=0
Horizontal Shear Equilibrium
𝑭 𝒙 = 𝟎 i.e.
Ha + Hd= 10kN

At Joint B Mba+Mbc=0
i.e. Equation (2)+ Equation(3)=0
𝟐𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆+𝟐𝑬𝑰𝜽 𝒃 − 𝑬𝑰𝜽 𝒄=0
𝟒𝑬𝑰𝜽 𝒃 − 𝑬𝑰𝜽 𝒄 −𝟏. 𝟓𝑬𝑰∆=0…………..(7)

At joint C Mcb+ Mcd=0
i.e. Equation (4)+ Equation(5)=0
𝑬𝑰𝜽 𝒃 + 𝟐𝑬𝑰𝜽 𝒄+𝟒𝑬𝑰𝜽 𝒄 − 𝟔𝑬𝑰∆=0
𝐄𝐈𝛉 𝐛+𝟔𝐄𝐈𝛉 𝐜- 𝟔𝐄𝐈∆=0…………..(8)

Horizontal Shear Equilibrium
𝑭 𝒙 = 𝟎 i.e.
Ha + Hd= 10kN
Ha=
𝑴𝒂𝒃+𝑴𝒃𝒂
𝟐
& Hd=
𝑴 𝒄𝒅
+𝑴𝒅𝒄
𝟏
(
𝟐
)+(
𝑴𝒄𝒅+𝑴𝒅𝒄
𝟏
)=10

Ha=
𝟐
=
𝑬𝒒 𝒏
(1)+𝑬𝒒 𝒏
(2)
𝟐
= 𝑬𝑰𝜽 𝒃−𝟏.𝟓𝑬𝑰∆ +𝟐𝑬𝑰𝜽 𝒃−𝟏.𝟓𝑬𝑰∆
𝟐
= 𝟏. 𝟓𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆
Hd=
𝑴𝒄𝒅+𝑴𝒅𝒄
𝟏
=
𝑬𝒒 𝒏
(5)+𝑬𝒒 𝒏
(6)
𝟏
= 4𝐸𝐼𝜽 𝒄−𝟔𝑬𝑰∆+2𝐸𝐼𝜽 𝒄−𝟔𝑬𝑰∆
𝟏
=6𝑬𝑰𝜽 𝒄 − 𝟏𝟐𝑬𝑰∆

Ha + Hd= 10kN i.e.
𝟏. 𝟓𝐄𝐈𝛉 𝐛
- 𝟏. 𝟓𝐄𝐈∆+ 𝟔𝐄𝐈𝛉 𝐜
- 𝟏𝟐𝐄𝐈∆ =10
1.5𝐄𝐈𝛉 𝐛+𝟔𝐄𝐈𝛉 𝐜- 13.5𝐄𝐈∆=10……(9)
Solving Eqn (7),(8) & (9) we get

𝟒𝑬𝑰𝜽 𝒃 − 𝑬𝑰𝜽 𝒄 − 𝟏. 𝟓𝑬𝑰∆=0 ……..(7)
𝐄𝐈𝛉 𝐛+𝟔𝐄𝐈𝛉 𝐜- 𝟔𝐄𝐈∆=0 ……..(8)
1.5𝐄𝐈𝛉 𝐛+𝟔𝐄𝐈𝛉 𝐜- 13.5𝐄𝐈∆=10 ……(9)
𝑬𝑰𝜽 𝒃= -0.833
𝑬𝑰𝜽 𝒄= -1.25
𝑬𝑰∆= −𝟏. 𝟑𝟖

Put the value of EIθb, EIθc & EI∆in Equation
(1), (2), (3) , (4),(5) &(6) to get final End
Moments,

𝑴 𝒂𝒃 = 𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆= 1.237kN.m
𝑴 𝒃𝒂 = 𝟐𝑬𝑰𝜽 𝒃 − 𝟏. 𝟓𝑬𝑰∆= 0.404kN.m
Mbc=𝟐𝑬𝑰𝜽 𝒃 − 𝑬𝑰𝜽 𝒄= -0.404kN.m
Mcb=𝑬𝑰𝜽 𝒃 + 𝟐𝑬𝑰𝜽 𝒄= -3.33kN.m
𝑴 𝒄𝒅 = 𝟒𝑬𝑰𝜽 𝒄 − 𝟔𝑬𝑰∆= 3.33kN.m
Mdc=𝟐𝑬𝑰𝜽 𝒄 − 𝟔𝑬𝑰∆= 5.78kN.m

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