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

Step 2 — Geometry of Deformation: Since the compound axial member is attached
to rigid supports at A and C, the overall elongation of the structure can be no more than
1 mm. In other words,
δ + δ = 1mm
(a)
1 2
Step 3 — Force–Temperature–Deformation Relationships: Write the force–
temperature–deformation relationships for the two members:
FL 1 1
FL 2 2
δ1
= + α1D TL1 and δ2
= + α2DTL2 (b)
AE
AE
1 1
2 2
Step 4 — compatibility Equation: Substitute Equations (b) into Equation (a) to
obtain the compatibility equation:
FL
AE
1 1
1 1
FL 2 2
+ α1D TL1
+ + α2D TL2 = 1mm
(c)
AE
2 2
Step 5 — Solve the Equations: Substitute F 2 = F 1 (from the equilibrium equation)
into Equation (c), and solve for the internal force F 1 :
⎡ L1
F1
⎢
⎣ AE
1 1
+
L2
AE
2 2
⎤
⎥ 1mm α1 TL1 α2 TL2
⎦
= − D − D (d)
In computing the value for F 1 , pay close attention to the units, making sure that they
are consistent:
⎡ 900 mm
600 mm ⎤
F1 ⎢
+
2 2 2 2 ⎥
⎣(2,000 mm )(70,000 N/mm ) (3,000 mm )(105,000 N/mm ) ⎦
−6 −6
= 1mm − (22.5 × 10 /C)(60 ° ° C)(900 mm) − (18.0 × 10 /C)(60 ° ° C)(600 mm)
Therefore,
The normal stress in rod (1) is
F1 =− 103,560 N =−103.6 kN
F1
103,560 N
s 1 = = − =− 51.8 MPa = 51.8 MPa ( C)
Ans.
2
A 2,000 mm
and the normal stress in rod (2) is
1
F2
103,560 N
s 2 = = − =− 34.5 MPa = 34.5 MPa ( C)
Ans.
2
A 3,000 mm
2
(e)
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