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

ExAmpLE 5.4
A tie rod (1) and a pipe strut (2) are used to support a 50 kN load,
as shown. The cross-sectional areas are A 1 = 650 mm 2 for tie rod
(1) and A 2 = 925 mm 2 for pipe strut (2). Both members are made
of structural steel that has an elastic modulus E = 200 GPa.
(a) Determine the normal stresses in tie rod (1) and pipe
strut (2).
(b) Determine the elongation or contraction of each member.
(c) Sketch a deformation diagram that shows the displaced
position of joint B.
(d) Compute the horizontal and vertical displacements of joint B.
y
A
(2)
1.25 m
(1)
B
50 kN
1.15 m
x
Plan the Solution
From a free-body diagram of joint B, the internal axial forces
in members (1) and (2) can be calculated. The elongation
(or contraction) of each member can then be computed from C
Equation (5.2). To determine the displaced position of joint B,
the following approach will be used: We will imagine that the pin at joint B is temporarily
removed, allowing members (1) and (2) to deform either in elongation or contraction.
Then, member (1) will be rotated about joint A, member (2) will be rotated about joint C,
and the intersection point of these two members will be located. We will imagine that the
pin at B is now reinserted in the joint at this location. The deformation diagram describing
the preceding movements will be used to compute the horizontal and vertical displacements
of joint B.
SOLUTION
(a) Member Stresses
The internal axial forces in members (1) and (2) can be determined from equilibrium
equations based on a free-body diagram of joint B. The sum of forces in the horizontal
(x) direction can be written as
F 1
(1)
B
Σ Fx = −F1 − F2cos 42.61°=
0
42.61°
and the sum of forces in the vertical (y) direction can be expressed as
(2)
Σ Fy = − F1
sin 42.61° − 50 kN = 0
∴ F
2
= −73.85 kN
Substituting this result into the previous equation gives
The normal stress in tie rod (1) is
F
1 =
54.36 kN
F1
(54.36 kN)(1, 000 N/kN)
2
s 1 = = = 83.63 N/mm (T) = 83.6 MPa(T) Ans.
2
A 650 mm
1
and the normal stress in pipe strut (2) is
F2
(73.85 kN)(1, 000 N/kN)
2
s 2 = = = 79.84 N/mm (C) = 79.8 MPa(C)
2
Ans.
A 925 mm
2
F 2
50 kN
99