find the deflection for B2 (point C TO F) using AMM (Area Moment Diagram) KGCA1.20 mHDB-4B-3B-26.3 mB-19.0 mE1.50 mL4.20 m4.20 mF4.20 mB Reactions:EM₁ = 0C10 x 0.6 + 26.82 × 1·2+29.25 × 7-56:3³)+ (16-25×9) (²1) - V₁ × 9 = 0V₁ = 143.92 KN+ 10 × 8·25+ (10·84 × 6·3) (1- 2 + 6≤ Fy= 0VVF-10-26.82-29-25-10- 10.84 x 6-3-V₂90.37= 146.69 KN16.25 x 9 = 0Calculations:=shear Force(just to the left = JLSE= 146.69 KNSFM (JL) = 146.69 -16.25 × 0·6 = 136.94 KNSFM (JR) = 146.69-10 · 16 · 25 × 0·6 = 126.94 KNSF₂ (JL) = 126-94-16.25x0.6 = 117.19 KNSF₂ (JR) = 117.19 - 26.82 = 90.37 KNKNSFE (JL) = 40.37 - 10.84 × 6.3 -16.25 x 6-3 = -80.3SEE (JR) = -80.3 - 29.25 = 109.55 KNSF (JL) = -109.55 - 16-25 x0.75=-121.74 KNSF (JR) = -121-74-10 = -131-74 KNSFF16.25 KNm80.36.3-4с= -131-74- 16.25 × 0·75 = -143.92 KN10 KN⇒x= 3·336mJust to the right = JR)|VcM D0.6m146.6926.82 KN1.2m136.94126.947117-19१०.37C M D85.09+158.33с mp10.84 KN/m(FBD309-1+(6.3-x)29-25KN80.3$109-55E6.3mof Beam B-2) VE10 KNN0.75mK-5mE N121.747131.74119011FE NFSFD(KN)143.92103.37BMD(KN-m)F

In case of Area moment method, First of all, draw the M/EI diagram of the beam. Then always try to…

Answered: find the deflection for B2 (point C TO…

Engineering

Civil Engineering

find the deflection for B2 (point C TO F) using AMM (Area Moment Diagram)

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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Question

find the deflection for B2 (point C TO F) using AMM (Area Moment Diagram)

K
G
C
A
1.20 m
H
D
B-4
B-3
B-2
6.3 m
B-1
9.0 m
E
1.50 m
L
4.20 m
4.20 m
F
4.20 m
B

Reactions:
EM₁ = 0
C
10 x 0.6 + 26.82 × 1·2+29.25 × 7-5
6:3³)
+ (16-25×9) (²1) - V₁ × 9 = 0
V₁ = 143.92 KN
+ 10 × 8·25+ (10·84 × 6·3) (1- 2 + 6
≤ Fy
= 0
VVF-10-26.82-29-25-10
- 10.84 x 6-3-
V₂
90.37
= 146.69 KN
16.25 x 9 = 0
Calculations:
=
shear Force
(just to the left = JL
SE
= 146.69 KN
SFM (JL) = 146.69 -16.25 × 0·6 = 136.94 KN
SFM (JR) = 146.69-10 · 16 · 25 × 0·6 = 126.94 KN
SF₂ (JL) = 126-94-16.25x0.6 = 117.19 KN
SF₂ (JR) = 117.19 - 26.82 = 90.37 KN
KN
SFE (JL) = 40.37 - 10.84 × 6.3 -16.25 x 6-3 = -80.3
SEE (JR) = -80.3 - 29.25 = 109.55 KN
SF (JL) = -109.55 - 16-25 x0.75=-121.74 KN
SF (JR) = -121-74-10 = -131-74 KN
SFF
16.25 KN
m
80.3
6.3-4
с
= -131-74- 16.25 × 0·75 = -143.92 KN
10 KN
⇒x= 3·336m
Just to the right = JR)|
Vc
M D
0.6m
146.69
26.82 KN
1.2m
136.94
126.94
7117-19
१०.37
C M D
85.09
+
158.33
с mp
10.84 KN/m
(FBD
309-1
+
(6.3-x)
29-25KN
80.3
$
109-55
E
6.3m
of Beam B-2) VE
10 KN
N
0.75m
K
-5m
E N
121.747
131.74
119011
F
E N
F
SFD
(KN)
143.92
103.37
BMD
(KN-m)
F

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