Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Solve for the equivalent force on an inclined surface. Look at Figure A.1 below. You will see a triangular block cut from a stress block with known stresses. The size and thickness of the block does not matter, so we will assume the area of the sloped inclined surface is A (sorry that it is the same symbol as the Point). If the area of the sloped surface is A, then the area of the other two surfaces is A cos (Theta) and A sin (Theta). First, draw a FBD of the block by multiplying each stress (normal and shear, known and unknown) by their associated area (in terms of A). Second, break the stress vectors on the side and bottom into components along the x’ and y’ axes. Third, sum the forces in the y’ direction and set them equal to zero (you should now be able to cross out the value of A for each term in your equation resulting in an equation with only one unknown, the normal stress in the y’ direction). Solve for the value of the normal stress in the y’ direction. Now repeat for equilibrium in the x’ direction to determine the shear on the inclined surface.

 
Beam with Three Point Loads
Stress Transformation
ELEVATION VIEW
y
18.75
X
x'
Theta 20 degrees
TA = 561 psi
ObA= 960 psi
A = the area of the sloped face of the block.
OyA
Think of a
triangular block
with the stresses
on each face.
TxYA
Theta = 20 degrees
ObA
TA = 561 psi
= 960 psi
expand button
Transcribed Image Text:Beam with Three Point Loads Stress Transformation ELEVATION VIEW y 18.75 X x' Theta 20 degrees TA = 561 psi ObA= 960 psi A = the area of the sloped face of the block. OyA Think of a triangular block with the stresses on each face. TxYA Theta = 20 degrees ObA TA = 561 psi = 960 psi
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