A sign of weight W is supported at its base by four bolls anchored in a concrete footing. Wind pressure P acts normal to the surface of the sign; the resultant of the uniform wind pressure is force fat the center of pressure (C.P). The wind force is assumed to create equal shear forces F/4 in the y direction at each boll (see figure parts a and c). The overturning effect of the wind force also causes an uplift force R at bolts A and C and a downward force (— R) al bolts B and D (see figure part b). The resulting effects of the wind and the associated ultimate stresses for each stress condition are normal stress in each boll ( h — 60 ksi); shear through the base plate (t h = 17 ksi); horizontal shear and bearing on each bolt ( t f u r = 25 ksi and cr^ = 75 ksi): and bearing on the bottom washer at B (or D) (a b o r = 50 ksi).
A sign of weight W is supported at its base by four bolls anchored in a concrete footing. Wind pressure P acts normal to the surface of the sign; the resultant of the uniform wind pressure is force fat the center of pressure (C.P). The wind force is assumed to create equal shear forces F/4 in the y direction at each boll (see figure parts a and c). The overturning effect of the wind force also causes an uplift force R at bolts A and C and a downward force (— R) al bolts B and D (see figure part b). The resulting effects of the wind and the associated ultimate stresses for each stress condition are normal stress in each boll ( h — 60 ksi); shear through the base plate (t h = 17 ksi); horizontal shear and bearing on each bolt ( t f u r = 25 ksi and cr^ = 75 ksi): and bearing on the bottom washer at B (or D) (a b o r = 50 ksi).
A sign of weight W is supported at its base by four bolls anchored in a concrete footing. Wind pressure P acts normal to the surface of the sign; the resultant of the uniform wind pressure is force fat the center of pressure (C.P). The wind force is assumed to create equal shear forces F/4 in the y direction at each boll (see figure parts a and c). The overturning effect of the wind force also causes an uplift force R at bolts A and C
and a downward force (— R) al bolts B and D (see figure part b). The resulting effects of the wind and the associated ultimate stresses for each stress condition are normal stress in each boll (
h — 60 ksi); shear through the base plate (t
h = 17 ksi); horizontal shear and bearing on each bolt ( t
fur = 25 ksi and cr^ = 75 ksi): and bearing on the bottom washer at B (or D) (a
bor = 50 ksi).
the bent bar AB weighing 10lb/ft is mounted as shown in figure 2 upon a carriage weighing 250 lb. The center of gravity of the carriage is at C midway between the wheels. If P=108lb and there is no frictional resistance at the wheels, find the wheel reaction R1. find the vertical component of the hinge force at A. find the horizontal concept of the hinge force at A.
A weight W = 300# is suspended by a cable
system as shown in the figure.
a. Calculate the moment of force W with
respect to point D (that means with respect
to an axis normal to the paper that goes
through D).
b. Determine all cable forces and the force in
post BE.
W=300 #
Hint: Draw two different FBD’s for points B
and C, and solve separately.
What is the magnitude of the resultant of tensions in cables BC and CD? Can you figure
this out without any calculation?
07
A
e)
As shown, an L-shaped bar is supported by a pin at joint A. The bar's dimensions are aaa = 620 mmmm and bbb = 400 mmmm , and the bar is subjected to a force with magnitude FFF = 5.05 kNkN at joint B. Ignoring the bar's weight, find the actual orientation of the applied force. What is the value of the angle θθtheta?
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EVERYTHING on Axial Loading Normal Stress in 10 MINUTES - Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=jQ-fNqZWrNg;License: Standard YouTube License, CC-BY