(1)Changethedistributedload from 400 lb/ft to 550 lb/ft.Keep inmind that this change will result in many other changes across the board
(1)Changethedistributedload from 400 lb/ft to 550 lb/ft.Keep inmind that this change will result in many other changes across the board
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
Page 425, Concept Problem 6.2. Solve this problem completely,butwith the following changes (Note that itinvolves the Sample Problem 5.7 on Page 376) (1)Changethedistributedload from 400 lb/ft to 550 lb/ft.Keep inmind that this change will result in many other changes across the board.Discuss all practical issues in your findings.Important: you must also show all your new intermediate results from Sample Problem 5.7.

Transcribed Image Text:A
V
A
-0.65
kips
3.2 kips
8 ft
(-18)
B
B
B
4.5 kips
4 ft →
(+18)
-3.85 kips
4.50
kips
C
X
Concept Application 6.2
Knowing that the allowable shearing stress for the timber beam of
Sample Prob. 5.7 is Tall = 0.250 ksi, check that the design is acceptable
from the point of view of the shearing stresses.
Recall from the shear diagram of Sample Prob. 5.7 that
Vmax = 4.50 kips. The actual width of the beam was given as
b = 3.5 in., and the value obtained for its depth was h : 14.55 in.
Using Eq. (6.10) for the maximum shearing stress in a narrow rec-
tangular beam,
=
Tmax
=
3 V 3 V
=
2 A
2 bh
3(4.50 kips)
2(3.5 in.) (14.55 in.)
= 0.1325 ksi
Since Tmax < Tall, the design obtained in Sample Prob. 5.7 is
acceptable.
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