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

Substitute these values into Equation (a), and derive the beam deflection at B:
v
B
PL ( /2)( L)
PL 11
=− [(2 L) − ( L/2) − ( L) ] =−
⎡
L
6(2 LEI )
24EI
⎣⎢ 4
2 2 2 2
3
⎤ 11PL
=− (b)
⎦⎥ 96EI
Case 2—Simply Supported Beam with unknown Reaction Force at B
Consider the simply supported beam AD with the unknown roller reaction applied as a
concentrated load at B. From Appendix C, the maximum deflection for a simply supported
beam with a concentrated load at midspan is given as
A
v
L
B
B y
v B
L
D
x
v
max
For this beam, let
L = 2L
EI
3
PL
=− (c)
48EI
P =−B (negative, since B acts upward)
y
= constant
Substitute these values into Equation (c) to obtain the following
expression for the beam deflection at B:
y
v
B
3 3
( By)(2 L)
BL y
=− − = (d)
48EI
6EI
Compatibility Equation
Add Equations (b) and (d) to obtain an expression for the beam deflection at B. Since B is
a roller support, the deflection at this location must be zero. Hence,
v
B
3 3
11PL
BL y
=− + = 0
96EI
6EI
(e)
As in the previous example, EI appears in both terms. Again, in other words, the specific
value of EI has no effect on the magnitude of the reaction for the roller support at B. From
the compatibility equation, the unknown roller reaction force can be expressed as
11
By = P
Ans.
16
ExAmpLE 11.7
A
v
L
B
L—
2
P
C
L—
2
D
x
Consider the beam and loading from Example 11.6. The beam
consists of a structural steel W530 × 66 wide-flange shape [E =
200 GPa; I = 351 × 10 6 mm 4 ]. Assume that P = 240 kN and
L = 5 m. Determine
(a) the reaction force at roller support B.
(b) the reaction force at B if the roller support settles 5 mm.
Plan the Solution
To answer part (a) of this problem, the equation developed for B y
in Example 11.6 will be used to calculate the reaction force. In
part (b) of this example, the middle roller settles 5 mm, meaning
466