STATICS / STRENGTH OF MATERIALS -
Example
In the structure shown below members ABC and
CDE are assumed to be solid rigid members. Member ABC is pinned
to the wall at A and is supported by a roller at point C. Member
CDE is pinned to the wall at point E, and is supported by steel
cable DF. Cable DF has a diameter of .75 inch. For this
structure:
A. Draw a Free Body Diagram showing all support forces
and loads.
B. Determine the axial stress in cable DF.
C. Determine the movement of point B due to the applied
load.
Unless otherwise indicated, all joints and support points are
assumed to be pinned or hinged joints.
Solution:
Part A.
STEP 1: Draw a free body diagram showing and labeling
all load forces and support (reaction) forces, as well as any
needed angles and dimensions.
STEP 2: Break any
forces not already in x and y direction into their x and y
components.
STEP 3: If we were to go about this problem in the
normal fashion, we would have five unknowns and only three
equations. There would be no way to solve this problem using that
approach. Therefore, we must take a different perspective. By
taking the member apart and analyzing member ABC there will only
be three unknowns.
Apply the equilibrium conditions:
Sum Fx = Ax = 0
Sum Fy = Ay + Cy - 12,000 lbs =
0
Sum TA = (-12,000 lbs)(6 ft) + Cy (8 ft) =
0
Solving for the unknowns:
Cy
= 9,000 lbs; Ay =
3,000 lbs
Now we can analyze member CDE in a similar fashion.
Apply the equilibrium conditions:
Sum Fx = Ex = 0
Sum Fy = Fy + Ey - 9,000 lbs = 0
Sum TE = -Fy (2 ft) + (9,000 lbs)(4 ft) = 0
Solving for the unknowns:
Fy
= 18,000 lbs; Ey =
-9,000 lbs
Part B. StressDF
= F/A = 18,000 lbs/ .4418 in2
= 40,740 psi
Part C. Def = Deformation
STEP 1: Point C moves due to
the deformation of cable FD. So we first determine the
deformation of FD.
DefFD = (FL / EA) = (18,000 lbs)(6 ft)(12 in/ft) / (30
*106)(.4417 in2) = .0978 in
STEP 2: Point C moves in proportion to
how much point D moves (see diagram).
Mov. C / 4 ft = Mov. D / 2 ft
Mov. C / 4 ft = .0978 in/ 2 ft
Mov. C = .1956 in
STEP 3: Member ABC is resting
on member CDE therefore, at point C the movement is the same for
members ABC and CDE (see diagram).
Mov. B / 6 ft = Mov. C / 8 ft
Mov. B / 6 ft = .1956 in/ 8 ft
Mov. B = .1467 in
Select:
Topic 3: Stress, Strain
& Hooke's Law - Table of Contents
Statics & Strength of
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