Page 377 - Marks Calculation for Machine Design

P1: Sanjay
15:14
January 4, 2005
Brown.cls
Brown˙C08
U.S. Customary MACHINE ASSEMBLY SI/Metric 359
Step 2. Substitute the given information in Step 2. Substitute the given information in
Eq. (8.81) to determine the moment of inertia Eq. (8.81) to determine the moment of inertia
(I group ) as (I group ) as
3 3
LH LH
I group = 2 + LHd 2 o I group = 2 + LHd 2 o
12 12
(2.5in)(0.265 in) (0.06 m)(0.007 m)
 3   3 
12 12
= 2   = 2  
+(2.5in)(0.265 in)(0.75 in) 2 +(0.06 m)(0.007 m)(0.02 m) 2
4
4
= 2((3.88 × 10 −3 + 3.727 × 10 −1 ) in ) = 2((1.7 × 10 −9 + 1.68 × 10 −7 ) m )
= 7.53 × 10 −1 in 4 = 3.39 × 10 −7 m 4
Step 3. Substitute the moment of inertia Step 3. Substitute the moment of inertia
(I group ) found in step 2 and the given informa- (I group ) found in step 2 and the given informa-
tion in Eq. (8.80) to determine (σ bending ) as tion in Eq. (8.80) to determine (σ bending ) as
PL o d o PL o d o
σ bending = σ bending =
I group I group
(800 lb)(18 in)(0.75 in) (3,600 N)(0.45 m)(0.02 m)
= 4 = −7 4
7.53 × 10 −1 in 3.39 × 10 m
10,800 lb · in 2 32.4N · m 2
= = −7 4
7.53 × 10 −1 in 4 3.39 × 10 m
= 14,343 lb/in 2 = 95,580,000 N/m 2
= 14.3 kpsi = 95.6MPa
= σ xx = σ xx
σ yy = 0 σ yy = 0
Step 4. Substitute the normal stresses Step 4. Substitute the normal stresses
(σ bending = σ xx ) and (σ yy = 0) from step 3 (σ bending = σ xx ) and (σ yy = 0) from step 3
in Eq. (5.14) to determine the average stress in Eq. (5.14) to determine the average stress
(σ avg ) as (σ avg ) as
σ xx + σ yy (14.3 kpsi) + (0) σ xx + σ yy (95.6MPa) + (0)
σ avg = = σ avg = =
2 2 2 2
= 7.15 kpsi = 47.8MPa
Step 5. Substitute the normal stresses Step 5. Substitute the normal stresses
(σ bending = σ xx ) and (σ yy = 0) from step 3 (σ bending = σ xx ) and (σ yy = 0) from step 3
and the shear stress (τ shear = τ xy ) from step 1 and the shear stress (τ shear = τ xy ) from step 1
in Eq. (5.14) to determine the maximum shear in Eq. (5.14) to determine the maximum shear
stress (τ max ) as stress (τ max ) as

2 2

σ xx − σ yy σ xx − σ yy
2
2
τ max = + τ xy τ max = + τ xy
2 2

2 2

(14.3) − (0) (95.6) − (0)
2
2
= + (0.6 ) kpsi = + (4.3 ) MPa
2 2

= (51.12) + (0.36 ) kpsi = (2,284.8) + (18.5 ) MPa
√
= 51.48 kpsi = 7.18 kpsi = 2,303.3 kpsi = 48.0MPa