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

that the roller support is displaced 5 mm downward. The compatibility equation developed in
Example 11.6 assumed that the beam deflection at B was zero. In this instance, however, the
compatibility equation must be revised to account for the 5 mm downward displacement.
SolutioN
(a) From Example 11.6, the reaction force at B for this beam and loading configuration
is given by
11
By
= P
16
Since P = 240 kN, the reaction force at B is B y = 165 kN.
(b) The compatibility equation derived in Example 11.6 was
v
B
3 3
11PL
BL y
=− + = 0
96EI
6EI
This equation was based on the assumption that the beam deflection at roller support B
would be zero. In part (b) of this example, however, the possibility that the support settles
by 5 mm is being investigated. That possibility is a very practical consideration. All building
structures rest on foundations. If these foundations are constructed on solid rock,
there may be little or no settlement; however, foundations that rest on soil or sand will
always settle to some extent. If all supports settle by the same amount, the structure will
be displaced as a rigid body and there will be no effect on the internal forces and moments
of the structure. However, if one support settles more that the others, then the reactions
and internal forces in the structure will be affected. Part (b) of this example examines the
change in reaction forces that would occur if the roller support at B were displaced downward
5 mm more than the displacements of supports A and C. This situation is termed
differential settlement.
So, roller support B settles 5 mm. The beam is connected to this support; therefore,
the beam deflection at B must be v B = –5 mm. The compatibility equation from Example
11.6 will be revised to account for this nonzero beam deflection at B, giving
v
B
3 3
11PL
BL y
=− + = −5mm
96EI
6EI
and an expression for the reaction force at B can be derived:
B
6EI
⎡
PL ⎤
= ⎢− 5mm +
11 3
L
⎥
⎣ 96EI
⎦
y 3
(a)
Unlike previous examples, EI does not cancel out of this equation. In other words, the
magnitude of B y will depend not only on the magnitude of the support settlement, but also
on the flexural properties of the beam. In Equation (a), the following values will be used:
P = 240 kN = 240,000 N
L = 5m=
5,000 mm
I
= 351 × 10 mm
6 4
E = 200 GPa = 200,000 MPa
Substitute these values into Equation (a), and compute B y . Pay particular attention to the
units associated with each variable, and make sure that the calculation is dimensionally
consistent. In this example, all force units will be converted to newtons and all length
units will be expressed in millimeters. When all is said and done, we have
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