Beam Displacements - Massachusetts Institute of … - formula to find displacement

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Beam Displacements
David Roylance
Department of Materials Science and Engineering
Massachusetts Institute of Technology
Cambridge, MA 02139
November 30, 2000
Introduction
We want to be able to predict the deflection of beams in bending, because many applications
have limitations on the amount of deflection that can be tolerated. Another common need for
deflection analysis arises from materials testing, in which the transverse deflection induced by a
bending load is measured. If we know the relation expected between the load and the deflection,
we can "back out" the material properties (specifically the modulus) from the measurement. We
will show, for instance, that the deflection at the midpoint of a beam subjected to "three-point
bending" (beam loaded at its center and simply supported at its edges) is
P L3
P = 48EI
where the length L and the moment of inertia I are geometrical parameters. If the ratio of P
to P is measured experimentally, the modulus E can be determined. A stiffness measured this
way is called the flexural modulus.
There are a number of approaches to the beam deflection problem, and many texts spend
a good deal of print on this subject. The following treatment outlines only a few of the more
straightforward methods, more with a goal of understanding the general concepts than with
developing a lot of facility for doing them manually. In practice, design engineers will usually
consult handbook tabulations of deflection formulas as needed, so even before the computer age
many of these methods were a bit academic.
Multiple integration
In Module 12, we saw how two integrations of the loading function q(x) produces first the shear
function V (x) and then the moment function M (x):
V = - q(x) dx + c1 (1)
M = - V (x) dx + c2 (2)
where the constants of integration c1 and c2 are evaluated from suitable boundary conditions on
V and M . (If singularity functions are used, the boundary conditions are included explicitly and
the integration constants c1 and c2 are identically zero.) From Eqn. 6 in Module 13, the curvature
1
v,xx(x) is just the moment divided by the section modulus EI. Another two integrations then
give
1
v,x(x) = EI M (x) dx + c3 (3)
v(x) = v,x(x) dx + c4 (4)
where c3 and c4 are determined from boundary conditions on slope or deflection.
Example 1
Figure 1: Three-point bending.
As an illustration of this process, consider the case of "three-point bending" shown in Fig. 1. This
geometry is often used in materials testing, as it avoids the need to clamp the specimen to the testing
apparatus. If the load P is applied at the midpoint, the reaction forces at A and B are equal to half the
applied load. The loading function is then
q(x) = P x -1 - P x - L
2 2 -1
Integrating according to the above scheme:
V (x) = - P x 0 + P x - L 0
22
M (x) = P x 1 - P x - L 1 (5)
22
EIv,x(x) = P x 2 - P x - L 2 + c3
4 22
From symmetry, the beam has zero slope at the midpoint. Hence v,x = 0 @ x = L/2, so c3 can be found
to be -P L2/16. Integrating again:
P x 3 - P x - L 3 - P L2x + c4
EIv(x) = 12 6 2 16
The deflection is zero at the left end, so c4 = 0. Rearranging, the beam deflection is given by
2
P 4x3 - 3L2x - 8 x - L 3 (6)
v = 48EI 2
The maximum deflection occurs at x = L/2, which we can evaluate just before the singularity term
activates. Then
P L3
max = 48EI (7)
This expression is much used in flexural testing, and is the example used to begin this module.
Before the loading function q(x) can be written, the reaction forces at the beam supports
must be determined. If the beam is statically determinate, as in the above example, this can
be done by invoking the equations of static equilibrium. Static determinacy means only two
reaction forces or moments can be present, since we have only a force balance in the direction
transverse to the beam axis and one moment equation available. A simply supported beam (one
resting on only two supports) or a simply cantilevered beam are examples of such determinate
beams; in the former case there is one reaction force at each support, and in the latter case there
is one transverse force and one moment at the clamped end.
Of course, there is no stringent engineering reason to limit the number of beam supports
to those sufficient for static equilibrium. Adding "extra" supports will limit deformations and
stresses, and this will often be worthwhile in spite of the extra construction expense. But the
analysis is now a bit more complicated, since not all of the unknown reactions can be found from
the equations of static equilibrium. In these statically indeterminate cases it will be necessary
to invoke geometrical constraints to develop enough equations to solve the problem.
This is done by writing the slope and deflection equations, carrying the unknown reaction
forces and moments as undetermined parameters. The slopes and deflections are then set to
their known values at the supports, and the resulting equations solved for the unknowns. If
for instance a beam is resting on three supports, there will be three unknown reaction forces,
and we will need a total of five equations: three for the unknown forces and two more for the
constants of integration that arise when the slope and deflection equations are written. Two
of these equations are given by static equilibrium, and three more are obtained by setting the
deflections at the supports to zero. The following example illustrates the procedure, which is
straightforward although tedious if done manually.
Example 2
Consider a triply-supported beam of length L = 15 as shown in Fig. 2, carrying a constant uniform load
of w = -10. There are not sufficient equilibrium equations to determine the reaction forces Ra, Rb, and
Rc, so these are left as unknowns while multiple integration is used to develop a deflection equation:
q(x) = Ra x -1 + Rb x - 7.5 -1 + Rc x - 15 -1 - 10 x 0
V (x) = - q(x) dx = -Ra x 0 - Rb x - 7.5 0 - Rc x - 15 0 + 10 x 1
M (x) = - V (x) dx = Ra x 1 + Rb x - 7.5 1 + Rc x - 15 1 - 10
2 x2
EIy (x) = M (x) dx = Ra x 2 + Rb x - 7.5 2 + Rc x - 15 2 - 10
2 2 2 6 x 3 + c1
3


What is the formula for calculating displacement?What is the formula for calculating displacement?If an object is moving with constant velocity, then.Displacement = velocity x time.If an object is moving with constant acceleration then the equation of third law of motion used to find displacement:S = ut + ½ at²S = v2−u22a.If v = final velocity,u = Initial velocity.s = displacement.

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