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

Now compute the deformations in each of the three members. Member (1) is a solid aluminum
rod 20 mm in diameter; therefore, its cross-sectional area is A 1 = 314.159 mm 2 , and
FL 1 1 (4 kN)(1, 000 N/kN)(2.0 m)(1,000 mm/m)
δ 1 = = = 0.364 mm
2
AE (314.159 mm )(70 GPa)(1,000 MPa/GPa)
1 1
Member (2) has a diameter of 24 mm; therefore, its cross-sectional area is
A 2 = 452.389 mm 2 , and we have
FL 2 2 ( 16kN)(1,000 N/kN)(2.5 m)(1,000 mm/m)
δ 2 = = − =−1.263 mm
2
AE (452.389 mm )(70 GPa)(1,000 MPa/GPa)
2 2
The negative value of δ 2 indicates that member (2) contracts.
Member (3) is a solid steel rod 16 mm in diameter. Its cross-sectional area is A 3 =
201.062 mm 2 , and
FL 3 3 (8 kN)(1, 000 N/kN)(3.0 m)(1,000 mm/m)
δ 3 = = = 0.597 mm
2
AE (201.062 mm )(200 GPa)(1,000 MPa/GPa)
3 3
Geometry of Deformations
Since the joint displacements of B, C, and D relative to joint A are desired, joint A will be
taken as the origin of the coordinate system. How are the joint displacements related to
the member deformations in the compound axial member? The deformation of an axial
member can be expressed as the difference between the displacements of the end joints of
the member. For example, the deformation of member (1) can be expressed as the difference
between the displacement of joint A (i.e., the −x end of the member) and the displacement
of joint B (i.e., the +x end of the member):
Similarly, for members (2) and (3),
δ = u − u
1
δ = u − u and δ = u − u
B
2 C B 3 D C
Since the displacements are measured relative to joint A, define the displacement of joint
A as u A = 0. The preceding equations can then be solved for the joint displacements in
terms of the member elongations:
u = δ , u = u + δ = δ + δ , and u = u + δ = δ + δ + δ
B 1 C B 2 1 2 D C 3 1 2 3
Using these expressions, we can now compute the joint displacements:
u = δ = 0.364 mm = 0.364 mm →
B 1
u = δ + δ = 0.364 mm + ( − 1.263 mm) =− 0.899 mm = 0.899 mm ←
C 2 2
u = δ + δ + δ = 0.364 mm + ( − 1.263 mm) + 0.597 mm = −0.302 mm
D 1 2 3
= 0.302 mm ← Ans.
A positive value for u indicates a displacement in the +x direction, and a negative u indicates
a displacement in the –x direction. Thus, joint D moves to the left even though tension
exists in member (3).
The nomenclature and sign conventions introduced in this example may seem
unnecessary for such a simple problem. However, the calculation procedure established
here will prove quite powerful as problems that are more complex are introduced, particularly
problems that cannot be solved with statics alone.
A
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