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

x
B
configurations due to a force applied at joint B. From the Pythagorean
theorem, the actual deformation in bar AB is
D
B
y
2 2
δ = ( L + y)
+ x − L
AB
Transposing the last term and squaring both sides gives
θ
δ
+ 2Lδ
+ L = L + 2Ly + y + x
2 2 2 2 2
AB AB
C
R
L
If the displacements are small (the usual case for stiff materials and
elastic action), the terms involving the squares of the displacements
may be neglected; hence, the deformation in bar AB is
A
δ AB
≈
y
FIGURE 5.9 Axial structure with intersecting
members.
The axial deformation of bar BC is
In a similar manner, the deformation in bar BD is
δ BD
≈ x
2 2
δ = ( R cos θ + x) + ( R sin θ + y)
− R
BC
Transposing the last term and squaring both sides gives
δ
+ 2Rδ
+ R = R cos θ + 2Rxcosθ + x + R sin θ + 2Rysinθ
+ y
2 2
BC BC
2 2 2 2 2 2
The second-degree displacement terms can be neglected since the displacements are small.
Using the trigonometric identity sin 2 θ + cos 2 θ = 1, we can express the deformation in
member BC as
δ BD
δ ≈ xcosθ + ysinθ
BC
θ
θ
B
or, in terms of the deformations of the other two bars,
δ ≈ δ cosθ + δ sinθ
BC BD AB
δ AB
B
δ BC
δ AB sin θ δ BD cos θ
FIGURE 5.10 Geometric interpretation of
member deformations.
The geometric interpretation of this equation is indicated by the shaded
right triangles in Figure 5.10.
The general conclusion that may be drawn from the preceding discussion
is that, for small displacements, the axial deformation in any bar may
be assumed to be equal to the component of the displacement of one end of
the bar (relative to the other end), taken in the direction of the unstrained
orientation of the bar. Rigid members of the system may change orientation
or position, but they will not be deformed in any manner. For example, if
bar BD of Figure 5.9 were rigid and subjected to a small upward rotation,
then point B could be assumed to be displaced vertically through a distance y,
and δ BC would be equal to y sin θ.
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