Page 179 - Marks Calculation for Machine Design

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January 4, 2005
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Brown.cls
Brown˙C04
COMBINED LOADINGS
element, but with the normal stress (σ yy ) equal to zero. Keep in mind that it is very rare to
have a completely general stress element in actual engineering practice. 161
Again, although it will be a review on the stress equations, consider the following exam-
ples to show how this combination of loadings result in quantitative information.
U.S. Customary SI/Metric
Example 2. Determine the stresses at the top Example 2. Determine the stresses at the top
of a rectangular beam, like the one in Fig. 4.10, of a rectangular beam, like the one in Fig. 4.10,
subjected to a combination of compressive axial subjected to a combination of compressive axial
and bending loads, where and bending loads, 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. Calculate the cross-sectional area (A) Step 1. Calculate the cross-sectional area (A)
of the beam. of the beam.
A = bh = (2.0in)(12.0in) = 24.0in 2 A = bh = (0.05 m)(0.3m) = 0.015 m 2
Step 2. Substitute this cross-sectional area and Step 2. Substitute this cross-sectional area and
the force (P) in the equation for compressive the force (P) in the equation for compressive
axial stress to give axial stress to give
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. Calculate the moment of inertia (I) for Step 3. Calculate the moment of inertia (I) for
the beam. the beam.
1 3 1 3 1 3 1 3
I = bh = (2.0in)(12.0in) I = bh = (0.05 m)(0.3m)
12 12 12 12
= 288 in 4 = 0.0001125 m 4
Step 4. Substitute this moment of inertia (I), Step 4. Substitute this moment of inertia (I),
the distance (y) to the top of the beam, and the distance (y) to the top of the beam, and
the bending moment (M) in the equation for the bending moment (M) in the equation for
maximumnegativenormalstressduetobending maximumnegativenormalstressduetobending
to give to give
My top My max
σ max =− σ max =−
I I
(96,000 in · lb)(6.0in) (12,000 N · m)(0.15 m)
=− =−
288 in 4 0.0001125 m 4
2
2
=− 2,000 lb/in =−2 kpsi =−16,000,000 N/m =−16 MPa
Step 5. Combine the two compressive stresses Step 5. Combine the two compressive stresses
foundinsteps2and4togiveamaximumnormal foundinsteps2and4togiveamaximumnormal
stress at the top (σ top ). stress at the top (σ top ).