Question 3.14: Beam deflection Figure 3.122 shows a simply supported beam, ...
Beam deflection
Figure 3.122 shows a simply supported beam, ABC, with a point load at the centre of its span.
Use strain energy to derive an expression for the central deflection of the beam (assuming bending strain energy only, no shear).
Reaction forces:
From symmetry,
Bending moment distribution:
Taking a section, X-X, in the part of the span, AB, and considering the left-hand part of the beam as a free body. Moments about X-X give,
∴ M = \frac{Wx}{2}
Strain energy:
The strain energy stored in AB is given by,
= \frac{W^{2}}{8EI} \left[\frac{x^{3}}{3} \right]^{L/2}_{0}
= \frac{W^{2}L^{3}}{192EI}
From symmetry,
U_{BC} = U_{AB}The total strain energy
U_{tot} = 2 U_{AB} = \frac{W^{2}L^{3}}{96EI}Deflection:
The central deflection, i.e. deflection at the load, is therefore given by,
\gamma _{B} = \frac{\partial U}{\partial W} = \frac{WL^{3}}{48EI}
Dummy load
A curved beam, whose geometric form is a quadrant in the vertical plane, as shown in Figure 3.123(a), has cross-sectional dimensions small compared with the quadrant radius.
Use strain energy to derive an expression for the horizontal and vertical
deflections of its tip due to a vertical load P.
Because of the slender nature of the curved beam, direct stress and transverse
shear strain energies may be neglected compared to bending strain energy.
To determine the horizontal deflection, a horizontal dummy load, Q, is added at the tip of the beam as shown in Figure 3.123(b).
Bending moment expression:
As also shown in Figure 3.123(b), to determine the bending moment distribution, a
section X-X is taken at position B, angle φ from the horizontal, and equilibrium of a
section AB of the beam is considered. Taking moments about X-X,
∴ M = – PR \cos \phi – QP (1 -\sin \phi ) (3.87)
Note that the bending moment is a function of both the applied load, P, and the dummy load, Q. Also, the sign of the bending moment is unimportant as it is squared in the strain energy integral (see below).
Strain energy:
The strain energy expression is now
Note that ds = Rdφ and that the limits of integration are 0 to π/2.
Vertical displacement:
The vertical displacement is now given by differentiating equation (3.88) with respect to the applied load P, as follows,
To simplify matters, the dummy load, Q, can be set to zero before the integration is
performed. We then have,
= \frac{PR^{3}}{EI} \int_{0}^{\pi /2}{ \frac{1}{2} (1+ \cos \phi) d\phi }
= \frac{PR^{3}}{EI} \left[\phi + \frac{1}{2} \sin 2\phi \right] _{0}^{\pi /2}
∴ u_{\nu } = \frac{\pi PR^{3}}{4EI}
Horizontal displacement:
The horizontal displacement is now given by differentiating equation (3.88) with respect to the dummy load, Q, as follows,
Again, setting the dummy load, Q, equal to zero before integrating we have,
u_{h} = \frac{PR^{3}}{EI} \int_{0}^{\pi /2}{ \cos \phi (1 – \sin \phi )]d\phi }
= \frac{PR^{3}}{2EI} \int_{0}^{\pi /2}{ \cos \phi – \frac{1}{2} \sin 2\phi d\phi }
= \frac{PR^{3}}{EI} \left[\sin \phi + \frac{1}{4} \cos 2\phi \right] _{0}^{\pi /2}
∴ u_{h} = \frac{PR^{3}}{2EI}
Note that the vertical deflection is π/2 × horizontal deflection.
Combined strain energy
A circular cross section, offset cantilever, ABC, as shown in Figure 3.124(a) is loaded at its free end by a load, P.
Neglecting strain energy due to bending shear, derive an expression for the vertical deflection at the load point.
In this example, section AB of the cantilever stores bending strain energy only, while section BC stores both bending and torsional strain energy.
Strain energy in AB:
Referring to Figure 3.124(b), which shows a section, X-X, taken distance x from A, the bending moment is obtained by considering equilibrium:
M + Px = 0 ∴ M = – Px
The strain energy, U_{AB}, is given by,
U_{AB} = \int{\frac{M^{2}}{2EI} ds }= \int_{0}^{L_{1}}{\frac{P^{2}x^{2}}{2EI} dx }
= \frac{P^{2}L_{1}^{3}}{6EI}
(note the limits 0 to L_1 for the length)
Strain energy in BC:
Referring to Figure 3.124(c), which shows a section, X-X, in BC, taken distance x from B, the bending moment and torque are obtained by considering equilibrium:
M + Px = 0 ∴ M = – Px
T + PL_{1} = 0 ∴ T = – PL_{1}The strain energy, U_{BC}, is now given by
U_{BC} = \int{\frac{M^{2}}{2EI} ds } + \int{\frac{T^{2}}{2GJ} ds }= \int_{0}^{L_{2}}{\frac{P^{2}x^{2}}{2EI} dx } + \int_{0}^{L_{2}}{\frac{P^{2}L^{2}_{1}}{2GJ} dx }
= \frac{P^{2}L_{2}^{3}}{6EI} + \frac{P^{2}L_{1}^{2}L_{2}}{2GJ}
So, the total strain energy in the cantilever is
U = U_{AB} + U_{BC} = P^{2} \left[\frac{L_{1}^{3} + L_{2}^{3}}{6EI} + \frac{L_{1}^{2} L_{2}}{2GJ} \right] (3.89)Deflection of the tip:
The deflection of the tip is now given by differentiating equation (3.89) with respect to the applied load, P, as follows,
