Page 181 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
14:25
Brown.cls
Brown˙C04
U.S. Customary COMBINED LOADINGS SI/Metric 163
Example 3. Determine the stresses on the ele- Example 3. Determine the stresses on the ele-
ment at the neutral axis of the rectangular beam ment at the neutral axis of the rectangular beam
of Example 2, where of Example 2, where
P = 4 kip = 6,000 lb P = 18 kN = 18,000 N
M = 8,000 ft · lb = 96,000 in · lb M = 12 kN · m = 12,000 N · m
V = 10,000 lb V = 45 kN = 45,000 N
h = 12.0 in h = 30.0 cm = 0.3 m
b = 2.0 in b = 5.0 cm = 0.05 m
solution solution
Step 1. The cross-sectional area (A) of the Step 1. The cross-sectional area (A) of the
beam was found in Example 2 to be beam was found in Example 2 to be
A = bh = (2.0in)(12.0in) = 24.0in 2 A = bh = (0.05 m)(0.3m) = 0.015 m 2
Step 2. Using this area (A) and the axial Step 2. Using this area (A) and the axial
force (P), the compressive stress was found in force (P), the compressive stress was found in
Example 2 to be Example 2 to be
P 6,000 lb P 18,000 N
σ =− =− σ =− =−
A 24.0in 2 A 0.015 m 2
2
2
=−250 lb/in =−0.25 kpsi =−1,200,000 N/m =−1.2MPa
Step 3. The moment of inertia (I) for the beam Step 3. The moment of inertia (I) for the beam
was found in Example 2 to be was found in Example 2 to be
1 1 1 1
3
3
I = bh = (2.0in)(12.0in) 3 I = bh = (0.05 m)(0.3m) 3
12 12 12 12
= 288 in 4 = 0.0001125 m 4
Step 4. The maximum first moment (Q max ) Step 4. The maximum first moment (Q max )
is needed, and is found for a rectangle using is needed, and is found for a rectangle using
Eq. (1.41) as Eq. (1.41) as
1 2 1 2 1 2 1 2
Q max = bh = (2.0in)(12.0in) Q max = bh = (0.05 m)(0.3m)
8 8 8 8
= 36 in 3 = 0.0005625 m 3
Step 5. Substitute the shear force (V ), the Step 5. Substitute the shear force (V ), the
maximum first moment (Q max ), the moment of maximum first moment (Q max ), the moment of
inertia (I), and the width (b) in Eq. (1.39) for inertia (I), and the width (b) in Eq. (1.39) for
the shear stress due to bending to give the shear stress due to bending to give
VQ max VQ max
τ max = τ max =
Ib Ib
3
3
(10,000 lb)(36 in ) (45,000 N)(0.0005625 m )
= =
4
4
(288 in )(2in) (0.0001125 m )(0.05 m)
360,000 lb · in 3 25.3125 N · m 3
= =
576 in 5 0.000005625 m 5
2
2
= 2,500 lb/in = 2.5 kpsi = 4,500,000 N/m = 4.5MPa