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

Equations (h) and (i) can be solved simultaneously in several ways. The hand solution
here will use the substitution method. First, solve Equation (i) for F 2 :
F
2
=
F 1(100 mm) − (6kN)(350 mm)
300 mm
(j)
Next, substitute this expression into Equation (h) and collect terms with F 1 on the
left-hand side of the equation:
FL 1 1
(100 mm) AE
1 1
[(100 mm/300 mm) F]
L
+
(300 mm) AE
2 2
1 2
1
1
350 mm L2
=− α1DTL1 − α2D TL2
+ (6 kN)
⎡ ⎤
100 mm 300 mm
⎣⎢ 300 mm ⎦⎥ (300 mm) AE
Simplifying and solving for F 1 then gives
⎡
500 mm
(1/3)(400 mm) ⎤
F1 ⎢
+
2 2 2 2
⎣(100 mm)(310 mm )(200,000 N/mm ) (300 mm)(620 mm )(70,000 N/mm )
⎥
⎦
Therefore,
1
−6
=− (11.9 × 10 /C)(20 ° ° C)(500 mm)
100 mm
1
−6
− (22.5 × 10 /C)(20 ° ° C)(400 mm)
300 mm
350 mm
400 mm
+ (6,000 N)
⎡ ⎤
2 2
⎣⎢ 300 mm ⎦⎥ (300 mm)(620 mm )(70,000 N/mm )
Substituting back into Equation (j) gives
F1 =− 17,328.8 N =− 17.33 kN = 17.33 kN (C)
F2 =− 12,776.3 N =− 12.78 kN = 12.78 kN (C)
The normal stresses in members (1) and (2) can now be determined:
2 2
F1
17,328.8 N
s1
= = − =− 55.9 MPa = 55.9 MPa (C)
2
A 310 mm
1
F2
12,776.3 N
s 2 = = − =− 20.6 MPa = 20.6 MPa (C)
2
A 620 mm
2
Ans.
Note: The deformation of member (1) can be computed as
FL 1 1
( −17,328.8 N)(500 mm)
δ1
= + α1D TL1 =
2 2
AE
(310 mm )(200,000 N/mm )
1 1
−6
+ (11.9 × 10 /C)(20 ° ° C)(500 mm)
=− 0.1397 mm + 0.1190 mm = −0.0207
mm
125