The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and no couples exerted on both A and B, determine (a) the tension in the cable, (b) the reactions at A and B.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at
A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and
no couples exerted on both A and B, determine (a) the tension in the cable, (b) the
reactions at A and B.
Z
D
12 in.
30 in.
H
4 in.
A
E
8 in.
B
F
4 in. 25 in.
20 in.
x
Transcribed Image Text:The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and no couples exerted on both A and B, determine (a) the tension in the cable, (b) the reactions at A and B. Z D 12 in. 30 in. H 4 in. A E 8 in. B F 4 in. 25 in. 20 in. x
• FBD analysis
1. There are in total
• Presenting all forces and couples using components:
1. Reacions on A: A =
2. Reactions on B: B =
3. Tension in cable EF:
o EF =
i+
• TEF=
TEFI +
4. Self weight of the plate W =
1. TAA =
2. [AB =
3. TAE =
4. TAW =
i+
1. MAA=
2. MAB=
3. ME=
4. MAG=
+1
+
i+
• Determination of arms for all forces, taking point A as rotating point:
i+
supports on the system, by assuming no couples on A and B, no axial trust on B, the reactions/forces on the FB are:
j+
j+
+!
i+
i+
| +
i+
j+
TEFI +
k (in);
k (in);
♦ j+
k;
• Determination of moments about A for each force:
• Equilibrium Equations:
1. ΣΜΑ = 0:
o Coefficient of i simplified to:
o Coefficient of j simplified to:
k (in);
k (in);
k (lb-in);
o Coefficient of k simplified to:
2. [F = 0:
o Coefficient of i simplified to:
o Coefficient of j simplified to:
o Coefficient of k simplified to:
TEFJ+
◆lb
TEFI +
k (lb-in);
◆k (lb);
◆ k (lb);
TEF +
TEF+
TEF+
TEF+
TEF +
TEF+
TEF K;
◆ k (lb-in);
TEFK (Ib-in);
◆
= 0, hence TEF=
By +
Bz = 0, hence Bz=
= 0, hence Ax =
+
◆ +
= 0, hence By =
lb;
+ +
lb;
= 0, hence Az =
lb;
lb;
= 0, hence Ay =
lb.
lb;
Transcribed Image Text:• FBD analysis 1. There are in total • Presenting all forces and couples using components: 1. Reacions on A: A = 2. Reactions on B: B = 3. Tension in cable EF: o EF = i+ • TEF= TEFI + 4. Self weight of the plate W = 1. TAA = 2. [AB = 3. TAE = 4. TAW = i+ 1. MAA= 2. MAB= 3. ME= 4. MAG= +1 + i+ • Determination of arms for all forces, taking point A as rotating point: i+ supports on the system, by assuming no couples on A and B, no axial trust on B, the reactions/forces on the FB are: j+ j+ +! i+ i+ | + i+ j+ TEFI + k (in); k (in); ♦ j+ k; • Determination of moments about A for each force: • Equilibrium Equations: 1. ΣΜΑ = 0: o Coefficient of i simplified to: o Coefficient of j simplified to: k (in); k (in); k (lb-in); o Coefficient of k simplified to: 2. [F = 0: o Coefficient of i simplified to: o Coefficient of j simplified to: o Coefficient of k simplified to: TEFJ+ ◆lb TEFI + k (lb-in); ◆k (lb); ◆ k (lb); TEF + TEF+ TEF+ TEF+ TEF + TEF+ TEF K; ◆ k (lb-in); TEFK (Ib-in); ◆ = 0, hence TEF= By + Bz = 0, hence Bz= = 0, hence Ax = + ◆ + = 0, hence By = lb; + + lb; = 0, hence Az = lb; lb; = 0, hence Ay = lb. lb;
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