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

mecmovies
ExAmpLE
m5.2 The roof and second floor of a building are supported
by the column shown. The structural steel [E = 200 GPa]
column has a constant cross-sectional area of 7,500 mm 2 .
Determine the deflection of joint C relative to foundation A.
ExAmpLE 5.2
A steel [E = 30,000 ksi] bar of rectangular cross section consists of a uniform-width
segment (1) and a tapered segment (2), as shown. The width of the tapered segment
varies linearly from 2 in. at the bottom to 5 in. at the top. The bar has a constant
thickness of 0.50 in. Determine the elongation of the bar resulting from application
of the 30 kip load. Neglect the weight of the bar.
C
5 in.
Plan the Solution
The elongation of uniform-width segment (1) may be determined from Equation
(5.2). The tapered segment (2) requires the use of Equation (5.5). An expression for
the varying cross-sectional area of segment (2) must be derived and used in the integral
for the 75 in. length of the tapered segment.
SOLUTION
For the uniform-width segment (1), the deformation from Equation (5.2) is
FL 1 1 (30 kips)(25 in.)
δ 1 = = = 0.0250 in.
AE (2 in.)(0.5 in.)(30,000 ksi)
1 1
75 in.
25 in.
B
A
(2)
y
x
2 in.
(1)
P = 30 kips
For tapered section (2), the width w of the bar varies linearly with position y. The
cross-sectional area in the tapered section can be expressed as
3in.
A2
( y) = wt =
⎡
2in. + ( y in.)
⎤
(0.5 in.) = 1 + 0.02y
in.
⎣⎢ 75 in. ⎦⎥
2
Since the weight of the bar is neglected, the force in the tapered segment is constant and
simply equal to the 30 kip applied load. Integrate Equation (5.5) to obtain
75 F2
A y E dy F2
75 1
( ) E A ( y)
dy 30 kips 75 1
δ 2 = ∫ =
30,000 ksi (1 0.02 y)
dy
0
∫ =
0
∫0
+
2 2
2 2
2 ⎛ 1 ⎞
= (0.001 in. ) [ln(1 0.02 y)] 0
75 0.0458 in.
⎝
⎜
0.02 in. ⎠
⎟ + =
91