Page 114 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
96
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
Step 3. As the distance (x) is between the Step 3. As the distance (x) is between the
supports, determine the bending moment from supports, determine the bending moment from
Eq. (2.55b)as Eq. (2.55b)as
w w
2
2
M = [L (x − a) − x ] M = [L (x − a) − x ]
2 2
15 lb/ft 225 N/m
= [(12 ft)(4ft − 2ft) = [(4m)(1.2m − 0.6m)
2 2
2
2
−(4ft) ] −(1.2m) ]
2
2
= (7.5 lb/ft) [(12)(2) − (16) ft ] = (112.5 N/m) [(4)(0.6) − (1.44) m ]
2
2
= (7.5 lb/ft) [(24 − 16) ft ] = (112.5 N/m) [(2.4 − 1.44) m ]
2
2
= (7.5 lb/ft)(8ft ) = 60 ft · lb = (112.5 N/m)(0.96 m ) = 108 N · m
Example 3. Calculate and locate the max- Example 3. Calculate and locate the max-
imum shear force (V max ) for the beam of imum shear force (V max ) for the beam of
Example 2, where Example 2, where
w = 15 lb/ft w = 225 N/m
L = 12 ft, a = 2ft L = 4m, a = 0.6 m
solution solution
The maximum shear force (V max ) occurs at the The maximum shear force (V max ) occurs at the
supports, given by Eq. (2.54). supports, given by Eq. (2.54).
w w
V max = (L − 2 a) V max = (L − 2 a)
2 2
15 lb/ft 225 N/m
= [(12 ft) − 2 (2ft)] = [(4m) − 2 (0.6m)]
2 2
= (7.5 lb/ft) [(12 − 4) ft] = (112.5 N/m) [(4 − 1.2) m]
= (7.5 lb/ft)(8ft) = 60 lb = (112.5 N/m)(2.8m) = 315 N
Example 4. Calculate and locate the maxi- Example 4. Calculate and locate the maxi-
mum bending moment (M max ) for the beam of mum bending moment (M max ) for the beam of
Example 3, where Example 3, where
w = 15 lb/ft w = 225 N/m
L = 12 ft, a = 2ft L = 4m, a = 0.6 m
solution solution
The maximum bending moment (M max ) occurs The maximum bending moment (M max ) occurs
either at the supports, given by Eq. (2.56), either at the supports, given by Eq. (2.56),
wa 2 (15 lb/ft)(2ft) 2 wa 2 (225 N/m)(0.6m) 2
M = = M = =
max @ 2 2 max @ 2 2
supports supports
2
2
(15 lb/ft)(4ft ) (225 N/m)(0.36 m )
= =
2 2
60 ft · lb 81 N · m
= = 30 ft · lb = = 40.5N · m
2 2