Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Uniformly distributed load between C and D: The uniformly distributed load requires the
use of two terms. From case 5 of Table 7.2, express this loading as
0 0
wx ( ) =−5kips/ft x − 12 ft + 5kips/ft x − 18 ft
(c)
Increasing linearly distributed load between C and D: Use case 6 of Table 7.2 to write the
following expression for the 6 kips/ft linearly distributed loading:
6kips/ft
1 6kips/ft
1 0
wx ( ) =− x − 12 ft + x − 18 ft + 6kips/ft x − 18 ft (d)
6ft
6ft
Reaction force E y : The upward reaction force at E is expressed by
1
wx ( ) = 19 kips〈 x − 24 ft〉 − (e)
As a practical matter, this term has no effect, since the value of Equation (e) is zero
for all values of x ≤ 24 ft. However, the term will be retained here for completeness
and clarity.
Complete-beam loading expression: The sum of Equations (a) through (e) gives the load
expression w(x) for the entire beam:
0
9kips/ft
1
9kips/ft
1
w( x) =− 9kips/ft 〈 x − 0ft〉 + 〈 x − 0ft〉 − 〈 x − 6ft〉
6ft
6ft
−
+ 56.0 kips〈 x − 6ft〉 − 5 kips/ft〈 x − 12ft〉 + 5 kips/ft〈 x − 18ft〉
6kips/ft
6kips/ft
− 〈 x − 12 ft〉 + 〈 x − 18 ft〉
6ft
6ft
+ 6kips/ft 〈 x − 18 ft〉 + 19 kips〈 x − 24 ft〉
1 0 0
1 1
0 −1
(f)
Shear-force equation: Integrate Equation (f), using the integration rules given in
Equation (7.11), to derive the shear-force equation for the beam:
∫
V( x) = w( x)
dx
9kips/ft
9kips/ft
=− 9kips/ft 〈 x − 0ft〉 + 〈 x − 0ft〉 − 〈 x − 6ft〉
2(6ft)
2(6ft)
1 2 2
+ 56.0 kips〈 x − 6ft〉 − 5 kips/ft〈 x − 12ft〉 + 5 kips/ft〈 x − 18ft〉
0 1 1
6kips/ft
6kips/ft
− 〈 x − 12 ft〉 + 〈 x − 18 ft〉
2(6ft)
2(6ft)
2 2
+ 6kips/ft 〈 x − 18 ft〉 + 19 kips〈 x − 24 ft〉
1 0
(g)
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