The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end is secured under the door knob and whose lower end is attached to a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the forces supported by hinges A and B when a horizontal force P=57 Ib is applied normal to the plane of the door as shown.

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end
a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the
forces supported by hinges A and B when a horizontal force P = 57 Ib is applied normal to the plane of the door as shown.
secured under the door knob and whose lower end is attached to
43"
P
46"
35"
8"
24"
Answers:
C =
1 Ib
Axy
-2 lb
%3D
Byy =
3 Ib
Transcribed Image Text:The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the forces supported by hinges A and B when a horizontal force P = 57 Ib is applied normal to the plane of the door as shown. secured under the door knob and whose lower end is attached to 43" P 46" 35" 8" 24" Answers: C = 1 Ib Axy -2 lb %3D Byy = 3 Ib
3/89
E Mno.
=0: 50 (48) - o.6 C (48) :
Ax
C = 100 Ib
P=
50 lb |
48'
60 lb
Ay
I Ma,=0:-50(8) - (0.6'100) 32
+64 By =0
By EMay-
B.
By = 36.25 lb
60 (4)-(0.8-10) 40
+ 64BX =0,
Bx = 27.5 lb
%3D
45.5 l6
=
%3D
EMB, =0:-50(72) + (0.6'100) (32) + 64Ay = 0
Ay
= 26.25 Ib
E MBy =0: 60(24) - 80 (48) +64 Ax=0
Ax = 27.5 1b
Axy=
= 38.0 Ib
Transcribed Image Text:3/89 E Mno. =0: 50 (48) - o.6 C (48) : Ax C = 100 Ib P= 50 lb | 48' 60 lb Ay I Ma,=0:-50(8) - (0.6'100) 32 +64 By =0 By EMay- B. By = 36.25 lb 60 (4)-(0.8-10) 40 + 64BX =0, Bx = 27.5 lb %3D 45.5 l6 = %3D EMB, =0:-50(72) + (0.6'100) (32) + 64Ay = 0 Ay = 26.25 Ib E MBy =0: 60(24) - 80 (48) +64 Ax=0 Ax = 27.5 1b Axy= = 38.0 Ib
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