Page 283 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
14:56
Brown˙C06
Brown.cls
STATIC DESIGN AND COLUMN BUCKLING
moment of inertia for a circle is given by Eq. (6.32) as
1 4 265
I circle = πr (6.32)
4
2
The area (A) of this circular cross section is (πr ), so the radius of gyration (k) for a
circle is given in Eq. (6.33) as

1 4
I circle 4 πr 1 2 r
k circle = = 2 = r = (6.33)
A circle πr 4 2
U.S. Customary SI/Metric
Example 2. Determine whether the following Example 2. Determine whether the follow-
circular column with diameter (d) and of length ing circular column, with diameter (d), and of
(L) with fixed–fixed ends is safe under a com- length (L) with fixed–fixed ends is safe under a
pressive axial force (P), where compressive axial force (P), where
P = 12,000 lb P = 54,000 N
L = 3ft=36in L = 1m
d = 2in d = 5cm = 0.05 m
3
3
E = 10 × 10 kpsi (aluminum) E = 70 × 10 MPa (aluminum)
S y = 40 kpsi S y = 270 MPa
C ends = 1.5 (fixed–fixed) adjusted C ends = 1.5 (fixed–fixed) adjusted
solution solution
Step 1. Using Eq. (6.33), calculate the radius Step 1. Using Eq. (6.33), calculate the radius
of gyration (k) as of gyration (k) as
r 1in r 0.025 m
k = = = 0.5in k = = = 0.0125 m
2 2 2 2
Step 2. Calculate the slenderness ratio (L/k). Step 2. Calculate the slenderness ratio (L/k).
L 36 in L 1m
= = 72 = = 80
k 0.5in k 0.0125 m
Step3. Calculatetheminimumslendernessra- Step3. Calculatetheminimumslendernessra-
tio(L/k) D fortheEulerformulafromEq.(6.29). tio(L/k) D fortheEulerformulafromEq.(6.29).
L 2π CE L 2π CE
2 1/2 2 1/2
= =
k k
D S y D S y
2 6 1/2 2 3 1/2
2π (1.5)(10 × 10 psi) 2π (1.5)(70 × 10 MPa)
= =
40,000 psi 270 MPa
6 1/2 4 1/2
296 × 10 psi 207.3 × 10 MPa
= =
40,000 psi 270 MPa
= (740) 1/2 = 86 = (7,676) 1/2 = 88