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

A
(1)
L 1+ v B
L1
100 mm
C
v B
B
(Note: Deflections shown
greatly exaggerated.)
F
By definition, the deformation in a member is the difference
between its final length (i.e., after the load is applied
and the temperature is increased) and its initial length.
For member (1), therefore,
δ = L − L = ( L + v ) − L = v
1 final initial 1 B 1
∴ v = δ
B
1
B
(c)
E
(2)
300 mm
D
Similarly, for member (2),
L 2
– v D
L 2
v D
6 kN
δ = L − L = ( L − v ) − L = −v
2 final initial 2 D 2
∴ v
D
= −δ
2
D
(d)
Substitute the results from Equations (c) and (d) into Equation
(b) to obtain
δ1 δ2
=− (e)
100 mm 300 mm
Step 3 — Force–Temperature–Deformation Relationships: Write the general
force–temperature–deformation relationships for the two axial members:
FL 1 1
FL 2 2
δ1
= + α1D TL1 and δ2
= + α 2 D TL 2
(f)
AE
AE
1 1
2 2
Step 4 — compatibility Equation: Substitute the force–temperature–deformation
relationships from Equation (f) into Equation (e) to obtain the compatibility equation:
1 ⎡ FL
100 mm
⎢
⎣ AE
1 1
1 1
⎤ 1 FL 2 2
+ α1DTL1⎥ α 2 TL 2
⎦
=− ⎡
300 mm AE
+ D ⎤
⎢
⎥
⎣ 2 2 ⎦
(g)
This equation is derived from information about the deflected position of the structure
and is expressed in terms of the two unknown member forces F 1 and F 2 .
Step 5 — Solve the Equations: In the compatibility equation [Equation (g)], group
the terms that include F 1 and F 2 on the left-hand side of the equation:
FL 1 1
FL 2 2
1
1
+ = − α1DTL1 − α2D TL2
(h)
(100 mm) AE (300 mm) AE 100 mm 300 mm
1 1
2 2
Equilibrium equation (a) can be treated in the same manner:
F(100 mm) − F (300 mm) = (6kN)(350 mm)
(i)
1 2
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