Page 286 - Marks Calculation for Machine Design

P1: Shibu
14:56
January 4, 2005
Brown.cls
Brown˙C06
STRENGTH OF MACHINES
268
was given as (40 kpsi) and the modulus of elasticity (E) was (10 × 10 kpsi). The distance
(c) for a circle is the radius (r), which in Example 2 is (1 in). 3
solution ec
Step 1. Calculate the eccentricity ratio as
k 2
ec (0.25 in)(1in)
= = 1
k 2 (0.5in) 2
Step 2. Substitute the eccentricity ratio found in step 1 and the known values of the other
terms in Eq. (6.10) to give Eq. (6.8) as
S y
σ cr =
ec 1 L σ cr

1 + s
k 2 2 k E
40 kpsi
= (6.35)
1 σ cr
1 + (1) s (72)
3
2 10 × 10 kpsi
40 kpsi
=
(72) √
1 + s σ cr
2(100)
40 kpsi
= √
1 + s (0.36) σ cr
where the units (kpsi) have been dropped for the modulus of elasticity (E) in the square
root term because the critical stress (σ cr ) will also have units of (kpsi). This is compatible
with the fact that the secant can only be evaluated for a nondimensional quantity.
As the critical stress (σ cr ) is on both sides of Eq. (6.8), it must be solved by trial-and-
error or some other numerical method. To show how quickly the trial-and-error method can
obtain a reasonably accurate value for the critical stress, start with an educated guess for
the critical stress, then modify this guess in successive iterations until the guess equals the
right hand side of Eq. (6.35). Stop when an appropriate level of accuracy is reached.
An excellent educated guess would be the yield stress divided by two, which would be
20 kpsi. Substitute the value 20 in the right hand side of Eq. (6.35) to give
40 kpsi
σ cr = √
1 + s[(0.36) σ cr ]
40 kpsi 40 kpsi 40 kpsi 40 kpsi
20 = √ = = =
1 + s[(0.36) 20] 1 + s [1.61] 1 + (−25.5) −24.5
=−1.6
As the right hand side came out negative, try a new guess of 10.
40 kpsi
σ cr = √
1 + s[(0.36) σ cr ]
40 kpsi 40 kpsi 40 kpsi 40 kpsi
10 = √ = = =
1 + s[(0.36) 10] 1 + s[1.14] 1 + (2.4) 3.4
= 11.8