5 m B 21 A -4 m- I с I D 200 KN 3 m E = constant = 70 GPa I = 1,290 (106) mm4
5 m B 21 A -4 m- I с I D 200 KN 3 m E = constant = 70 GPa I = 1,290 (106) mm4
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![Using Moment area method, calculate the horizontal deflection of Joint D in the
structure shown in Figure for Question 2 (a) below. I = 1290×106 mm².](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5334ecb1-0a8f-45e6-ad6b-e9d03822a8f8%2F19a71290-2dc0-4f07-b030-27eb7429c661%2Fy5ew17_processed.png&w=3840&q=75)
Transcribed Image Text:Using Moment area method, calculate the horizontal deflection of Joint D in the
structure shown in Figure for Question 2 (a) below. I = 1290×106 mm².
![5m
B
21
-4 m
I
с
I
D
200 KN
3 m
E = constant = 70 GPa
I = 1,290 (106) mm4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5334ecb1-0a8f-45e6-ad6b-e9d03822a8f8%2F19a71290-2dc0-4f07-b030-27eb7429c661%2Fd3pyu_processed.png&w=3840&q=75)
Transcribed Image Text:5m
B
21
-4 m
I
с
I
D
200 KN
3 m
E = constant = 70 GPa
I = 1,290 (106) mm4
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I cannot understand these
For point of contraflexural,
40090300=200×x90300
x=2 m
θA=0 (fixed support)
θDA=θD-θA=θD=Area of MEI Dia between A and D
= 19030012×600×3+600×4+12×300×3-12×200×2
=355090300
θD=0.03931 radianθD= 2.252°
Deflection at point D
∆D=31-CosθD
∆D=2.3176×10-3 mm
Deflection at point D is 2.3176×10-3 mm
Solution
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