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.

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Could you please solve the problem below, with work, in the same method as the solution posted below? Thank you!

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

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