Page 142 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C02
Brown.cls
124
U.S. Customary 16:18 STRENGTH OF MACHINES SI/Metric
Step 2. Determine the bending moment (M) Step 2. Determine the bending moment (M)
from Eq. (2.81). from Eq. (2.81).
wx 3 (300 lb/ft)(2ft) 3 wx 3 (4,500 N/m)(0.6m) 3
M =− =− M =− =−
6L 6 (6ft) 6 L 6 (1.8m)
2,400 ft · lb 972 N · m
=− =−
36 ft 10.8m
=−67 ft · lb =−90 N · m
Example 3. Calculate and locate the max- Example 3. Calculate and locate the max-
imum shear force (V max ) and the maximum imum shear force (V max ) and the maximum
bending moment (M max ) for the beam of bending moment (M max ) for the beam of
Examples 1 and 2, where Examples 1 and 2, where
w = 300 lb/ft w = 4,500 N/m
L = 6ft L = 1.8 m
solution solution
Step 1. Calculate the maximum shear force Step 1. Calculate the maximum shear force
(V max ) from Eq. (2.80) as (V max ) from Eq. (2.80) as
wL (300 lb/ft)(6ft) wL (4,500 N/m)(1.8m)
V max = = V max = =
2 2 2 2
1,800 lb 8,100 N
= = 900 lb = = 4,050 N
2 2
Step 2. Figure 2.115 shows that this maximum Step 2. Figure 2.115 shows that this maximum
shear force (V max ) of 900 lb occurs at the right shear force (V max ) of 4,050 N occurs at the right
end of the beam. end of the beam.
Step 3. Calculate the maximum bending Step 3. Calculate the maximum bending
moment (M max ) from Eq. (2.82) as moment (M max ) from Eq. (2.82) as
wL 2 (300 lb/ft)(6ft) 2 wL 2 (4,500 N/m)(1.8m) 2
M max = = M max = =
6 6 6 6
10,800 ft · lb 14,580 N · m
= =
6 6
= 1,800 ft · lb = 2,430 N · m
Step 4. Figure 2.116 shows that this maximum Step 4. Figure 2.116 shows that this maximum
bending moment (M max ) of 1,800 ft · lb occurs bending moment (M max ) of 2,430 N · m occurs
at the right end of the beam, meaning at the wall at the right end of the beam, meaning at the wall
support. support.

Deflection. For this loading configuration, the deflection ( ) along the beam is shown in
Fig. 2.117, and given by Eq. (2.83) for all values of the distance (x) from the left end of the
beam, as
w 5 4 5
= (x − 5 L x + 4 L ) 0 ≤ x ≤ L (2.83)
120 EIL
where = deflection of beam
w = triangular distributed load
x = distance from left end of beam