Read and Analyze the sample problems given. Give a thorough analysis stating what type of situation the author dealt, how the problem was solved, the assumptions that the author considered in solving the problem, and why where these assumptions made

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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Read and Analyze the sample problems given. Give a thorough analysis stating what type of situation the author dealt, how the problem was solved, the assumptions that the author considered in solving the problem, and why where these assumptions made.

Sample Problem 7.3
L.
Determine the maximum force P that can be applied to block A in Fig. (a) without
causing either block to move.
Surface
A
4, = 0.2
WA = 100 lb
Solution
B
The problem statement indicates that we are to find P that would cause impend-
ing motion. However, there are two possible ways in which motion can impend:
Surface e impending sliding at surface 1, or impending sliding at surface 2. Because
impending sliding is specified but not its location, this is a Type III problem.
The free-body diagrams of the entire system and each block are shown in
Figs. (b) and (c), respectively. Note that the equilibrium of each block yields
Wg = 200 lb
4, = 0.1
(a)
Transcribed Image Text:Sample Problem 7.3 L. Determine the maximum force P that can be applied to block A in Fig. (a) without causing either block to move. Surface A 4, = 0.2 WA = 100 lb Solution B The problem statement indicates that we are to find P that would cause impend- ing motion. However, there are two possible ways in which motion can impend: Surface e impending sliding at surface 1, or impending sliding at surface 2. Because impending sliding is specified but not its location, this is a Type III problem. The free-body diagrams of the entire system and each block are shown in Figs. (b) and (c), respectively. Note that the equilibrium of each block yields Wg = 200 lb 4, = 0.1 (a)
W = 100 lb
N = 100 lb
W = 100 lb
A
N = 100 lb
В
Wg = 200 lb
W = 200 lb
F2
IN = 300 Ib
F2
N2 = 300 lb
(b)
N = 100 lb and N, = 300 lb, as shown on the FBDS. Attention should be paid
to the friction forces. The friction force F2 on the bottom of block B is directed
to the left, opposite the direction in which sliding would impend. At surface 1,
block A would tend to slide to the right, across the top of block B. Therefore, F,
is directed to the left on block A, and to the right on block B. The tendency of F1
to slide B to the right is resisted by the friction force F2. Note that F1 and N1 do
not appear in the FBD in Fig. (b), because they are internal to the system of both
blocks.
Two solutions are presented here to illustrate both methods of analysis
described in Art. 7.3.
Method of Analysis 1
First, assume impending sliding at surface 1. Under this assumption we have
F1 = (Fi)max = (4,)¡ N1 = 0.2(100) = 20 lb
The FBD of block A then gives
EF, = 0 +,
P – F = 0
P = F = 20 lb
Next, assume impending sliding at surface 2, which gives
F2 = (F2)max = (14,)2 N2 = 0.1(300) = 30 lb
From the FBD of the entire system, Fig. (b), we then obtain
EF = 0
P - F2 = 0
P = F2 = 30 lb
Transcribed Image Text:W = 100 lb N = 100 lb W = 100 lb A N = 100 lb В Wg = 200 lb W = 200 lb F2 IN = 300 Ib F2 N2 = 300 lb (b) N = 100 lb and N, = 300 lb, as shown on the FBDS. Attention should be paid to the friction forces. The friction force F2 on the bottom of block B is directed to the left, opposite the direction in which sliding would impend. At surface 1, block A would tend to slide to the right, across the top of block B. Therefore, F, is directed to the left on block A, and to the right on block B. The tendency of F1 to slide B to the right is resisted by the friction force F2. Note that F1 and N1 do not appear in the FBD in Fig. (b), because they are internal to the system of both blocks. Two solutions are presented here to illustrate both methods of analysis described in Art. 7.3. Method of Analysis 1 First, assume impending sliding at surface 1. Under this assumption we have F1 = (Fi)max = (4,)¡ N1 = 0.2(100) = 20 lb The FBD of block A then gives EF, = 0 +, P – F = 0 P = F = 20 lb Next, assume impending sliding at surface 2, which gives F2 = (F2)max = (14,)2 N2 = 0.1(300) = 30 lb From the FBD of the entire system, Fig. (b), we then obtain EF = 0 P - F2 = 0 P = F2 = 30 lb
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