Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

y
A
2.4 m 1.8 m
Rigid bar
(1) (2) (3)
determine the final position of the rigid bar, we must first compute the forces in the three
axial members, using equilibrium equations. Then, Equation (5.2) can be used to compute
the deformation in each member. A deformation diagram can be drawn to define the relationships
between the rigid-bar deflections at A, B, and C. Then, the member deformations
will be related to those deflections. Finally, the deflection of joint D can be computed
from the sum of the rigid-bar deflection at C and the elongation in member (3).
B
C
x
SOLUTION
Equilibrium
Draw a free-body diagram (FBD) of the rigid
bar and write two equilibrium equations:
∑ Fy
=−F1 − F2 − F3
= 0
∑ M = (2.4 m) F − (1.8 m) F = 0
B
1 3
F 1 F 2 F 3
Force–Deformation Relationships
Compute the deformations in each of the three members:
By inspection, F 3 = P = 30 kN. Using this result,
we can simultaneously solve the two equations
to give F 1 = 22.5 kN and F 2 = –52.5 kN.
FL 1 1 (22.5 kN)(1, 000 N/kN)(3.6 m)(1,000 mm/m)
δ 1 = = = 9.00 mm
2
AE (500 mm )(18 GPa)(1,000 MPa/GPa)
1 1
FL 2 2 (
δ = = − 52.5 kN)(1, 000 N/kN)(3.6 m)(1,000 mm/m)
2
=−7.00 mm
2
AE (1,500 mm )(18 GPa)(1,000 MPa/GPa)
2 2
FL 3 3 (30 kN)(1,000 N/kN)(3.0 m)(1,000 mm/m)
δ 3 = = = 10.00 mm
2
AE (500 mm )(18 GPa)(1,000 MPa/GPa)
3 3
y
v A
Rigid bar B
C
v
A
B
v C
L 1
F
The negative value of δ 2 indicates that member (2) contracts.
2.4 m 1.8 m
L 1+ v A L2
E
L –
2 v B
D
x
Geometry of Deformations
Sketch the final deflected shape
of the rigid bar. Member (1) elongates,
so A will deflect upward.
Member (2) contracts, so B will
deflect downward. The deflection
of C must be determined.
(Note: Vertical deflections of
the rigid-bar joints are denoted by
v. In this example, the displacements
v A , v B , and v C are treated as
absolute values and the sketch is
used to establish the relationship
between the joint displacements.)
The rigid-bar deflections at
joints A, B, and C can be related
by similar triangles:
v + v v − v
=
2.4 m 1.8 m
A B C B
1.8 m
∴ v = v + v + v = v + v + v
2.4 m ( ) 0.75( )
C A B B A B B
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