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

Step 3 — Force–Deformation Relationships: We know that the elongation in an
axial member can be expressed by Equation (5.2). Therefore, the relationship between
the internal axial force in a member and the member’s deformation can be expressed
for each member as
δ
1
FL 1 1
= and δ2
=
AE
1 1
FL
AE
2 2
2 2
(d)
Step 4 — compatibility Equation: The force–deformation relationships [Equation
(d)] can be substituted into the geometry-of-deformation equation [Equation (c)] to
obtain a new equation based on deformations but expressed in terms of the unknown
member forces F 1 and F 2 :
FL 1 1
vA = vB = vC
∴ =
AE
1 1
FL
AE
2 2
2 2
(e)
Step 5 — Solve the Equations: From compatibility equation (e), derive an
expression for F 1 :
F
F L 2
2 A1
E1
(2 m) (550 mm ) (70 GPa)
= = F2
= 0.2139 F
2 2 (f)
L A E (2 m) (900 mm ) (200 GPa)
1 2
1
2
2
Substitute Equation (f) into Equation (a) and solve for F 1 and F 2 :
∑ F = 2F + F = 2(0.2139 F ) + F = 220 kN
y 1 2 2 2
∴ F = 154.083 kN and F = 32.958 kN
2 1
The normal stress in aluminum bars (1) is
and the normal stress in steel bar (2) is
F1
32,958 N
s 1 = = = 59.9 MPa ( T)
Ans.
2
A 550 mm
1
F2
154,083 N
s 2 = = = 171.2 MPa ( T)
Ans.
2
A 900 mm
2
From Equation (c), the deflection of the rigid beam is equal to the deformation of the
axial members. Since both members (1) and (2) elongate by the same amount, either
term in Equation (d) can be used. We thus have
FL 1 1 (32,958 N)(2,000 mm)
δ 1 = = = 1.712 mm
2 2
AE (550 mm )(70,000 N/mm )
1 1
Therefore, the rigid-beam deflection is v A = v B = v C = δ 1 = 1.712 mm.
By inspection, the rigid beam deflects downward.
Ans.
105