Deflection

Deflection and Stiffness

linear spring.

nonlinear stiffening spring.

nonlinear softening spring.

spring constant

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Tension, Compression, and Torsion a uniform bar in pure tension or compression, This equation does not apply to a long bar loaded in compression if there is a possibility of buckling The angular deflection of a uniform solid or hollow round bar subjected to a twisting moment T where θ is in radians.

where θ is in degrees.

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Deflection Due to Bending The curvature of a beam subjected to a bending moment M

where Ď is the radius of curvature.

Stress formula

For many problems in bending, the slope is very small, and for these the denominator of above Equation can be taken as unity successively differentiating this Equation Mohr’s 11/15/2013 yields circle

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EXAMPLE Using above Eqs., determine the equations for the slope and deflection of the beam, the slopes at the ends, and the maximum deflection.

Solution

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The boundary conditions for the simply supported beam are y = 0 at x = 0 and l.

Applying the first condition, y = 0 at x = 0, results in C2 = 0. Applying the second condition to Eq. with C2 = 0,

The maximum deflection occurs where dy/dx = 0. Substituting x = l/2

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Beam Deflections by Superposition A cantilever beam with an end load. Normally we model this problem by considering the left support as rigid. After testing the rigidity of the wall it was found that the translational stiffness of the wall was kt force per unit vertical deflection, and the rotational stiffness was kr moment per unit angular (radian) deflection. Determine the deflection equation for the beam under the load F.

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Here we will superpose the modes of deflection.

They are: (1)translation due to the compression of spring kt , (2) rotation of the spring kr , and (3) the elastic deformation of beam 1 given in Table A–9. The force in spring kt is R1 = F,

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Beam Deflections by Singularity Functions

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EXAMPLE 01 Figure shows the loading diagram for a beam cantilevered at A with a uniform load of 20 lbf/in acting on the portion 3 in ≤ x ≤ 7 in, and a concentrated counterclockwise moment of 240 lbf · in at x = 10 in. Derive the shear-force and bending moment relations, and the support reactions M1 The first two and R1.

integrations of q(x) for V(x) and M(x) do not require constants of integration provided all loads on the beam are accounted for in q(x).

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Figures b and c show the shearforce and bending-moment diagrams. Note that the impulse terms in Eq. (2), −M1<x>−1 and −240<x – 10>−1, are physically not forces and are not shown in the V diagram. Also note that both the M1 and 240 lbf . in moments are counterclockwise and negative singularity functions; however, by the convention the M1 and 240 lbf · in are negative and positive bending moments, respectively, 11/15/2013 which are reflected in Figures.

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EXAMPLE 02 A simply supported beam having a concentrated load F not in the center.

Develop the deflection equations using singularity functions.

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Integrating Eqs.

as indefinite integrals gives

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1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 1111111111111111111111111 111111111111111111111

EXAMPLE 03 Determine the deflection equation for the simply supported beam with the load distribution shown in Figure.

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EXAMPLE 04 The steel step shaft shown in Figure is mounted in bearings at A and F. A pulley is centered at C where a total radial force of 600 lbf is applied. Using singularity functions evaluate the shaft displacements at 1/2-in increments. Assume the shaft is simply supported.

Solution The reactions are found to be R1 = 360 lbf and R2 = 240 lbf.

Ignoring R2, using singularity functions, the moment equation is For simplification, we will consider only the step at D. That is, we will assume section AB has the same diameter as BC and section EF has the same diameter as DE. Since these sections are short and at the supports, the size reduction will not add much to the deformation.

The second area moments for BC and DE are

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A plot of M/I is shown. The values at points b and c, and the step change are

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Where x and y are in inches. We see that the greatest deflection is at x = 8.5 in, where y = −0.009385 in. Substituting C1 into Eq. (3) the slopes at the supports are found to be θA = 1.686(10−3) rad = 0.09657 deg, and θF = 1.198(10−3) rad = 0.06864 deg.

You might think these to be insignificant deflections, but as you will see in next chapters, on shafts, they are not. A Finite-element analysis was performed for the same model and resulted in y|x=8.5 in = −0.009380 in, θA = −0.09653◦ , θF = 0.06868◦ Virtually the same answer save some round-off error in the equations. If the steps of the bearings were incorporated into the model, more equations result, but the process is the same. The solution to this model is y|x=8.5 in = −0.009387 in, θA = −0.09763◦ , θF = 0.06973◦ The largest difference between the models is of the order of 1.5 percent.

Thus the simplification was justified.

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Bending

,

, For small deflections, ds â&#x2030;&#x2C6; dx.

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Transverse Shear An approximate solution can be obtained with a correction factor whose value depends upon the shape of the cross section. If we use C for the correction factor and V for the shear force, then the strain energy due to shear in bending is

A cantilever beam with a round cross section has a concentrated load F at the end, as shown. Find the strain energy in the beam

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Castiglianoâ&#x20AC;&#x2122;s theorem states that when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force.

,

Force and displacement in this statement are broadly interpreted to apply equally to moments and angular displacements The relative contribution of transverse shear to beam deflection decreases as the length-to-height ratio of the beam increases, and is generally considered negligible for l/d > 10.

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Castiglianoâ&#x20AC;&#x2122;s theorem can be used to find the deflection at a point even though no force or moment acts there. The procedure is: 1) Set up the equation for the total strain energy U by including the energy due to a fictitious force or moment Q acting at the point whose deflection is to be found.

2) Find an expression for the desired deflection δ, in the direction of Q, by taking the derivative of the total strain energy with respect to Q. 3) Since Q is a fictitious force, solve the expression obtained in step 2 by setting Q equal to zero. Thus, the displacement at the point of application of the fictitious force Q is In cases where integration is necessary to obtain the strain energy, it is more efficient to obtain the deflection directly without explicitly finding the strain energy, by moving the partial derivative inside the integral.

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This method is especially helpful if the force is a fictitious force Q, since it can be set to zero as soon as the derivative is taken.

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EXAMPLE 05

Using Castiglianoâ&#x20AC;&#x2122;s method, determine the deflections of points A and B due to the force F applied at the end of the step shaft shown. The second area moments for sections AB and BC are I1 and 2I1, respectively.

The deflection at A

For B, a fictitious force Q is necessary at the point. Assuming Q acts down at B, and x is as before, the moment equation is 11/15/2013

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Once the derivative is taken, Q can be set to zero

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EXAMPLE 06 For the wire form of diameter d shown in Figure, determine the deflection of point B in the direction of the applied force F (neglect the effect of transverse shear).

Element BC is in bending only

Element CD is in bending and in torsion.

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Member DG is axially loaded and is bending in two planes. The axial loading is constant

where l is the length of the member.

Thus, for the axial loading of DG, Fi = F, l = c, and

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With MDG2 = Fb and MDG1 = Fa.

Adding all deflections and noting that I = πd4/64, J = 2I, A = πd2/4, and G = E/[2(1 + ν)], we find that the deflection of B in the direction of F is

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EXAMPLE 07 The cantilevered hook shown is formed from a round steel wire with a diameter of 2 mm. The hook dimensions are l = 40 and R = 50 mm. A force P of 1 N is applied at point C. Use Castiglianoâ&#x20AC;&#x2122;s theorem to estimate the deflection at point D at the tip.

Since l/d and R/d are significantly greater than 10, only the contributions due to bending will be considered. To obtain the vertical deflection at D, a fictitious force Q will be applied there.

The normal and shear forces, N and V respectively, are shown but are considered negligible in the deflection analysis.

Since the derivative with respect to Q has been taken, we can set Q equal to zero, inserting Eqs. (1) and (2), 11/15/2013

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For section BC, with the variable of integration θ, summing moments about the break gives the moment equation for section BC.

inserting Eqs. (4) and (5) and setting Q = 0, we get

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Noting that the break in section CD contains nothing but Q, and after setting Q = 0, we can conclude that there is no actual strain energy contribution in this section.

Combining terms from Eqs. (3) and (6) to get the total vertical deflection at D,

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EXAMPLE 08 Given as shown , find the deflection at point 1

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Angular deflection at wall is zero

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EXAMPLE 09

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Since the shaded area showed failures in actual tests Most designers select point T such that Pcr/A = Sy/2

The parabolic curve is

then a = Sy

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The solution for the boundary conditions that y = 0 at x = 0, l is

By substituting x = l/2

ec/k2

=

eccentricity ratio.

Maximum bending moment also occurs at midspan

Substituting Mmax By imposing the compressive yield strength Syc as the maximum value of ﾏツ 11/15/2013

Secant Column Formula.

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Magnitude of the maximum compressive stress

where k = (I/A)1/2 and is the radius of gyration, This difference between the two formulas suggests that one way of differentiating between a â&#x20AC;&#x153;secant columnâ&#x20AC;? and a strut, or short compression member, is to say that in a strut, the effect of bending deflection must be limited to a certain small percentage of the eccentricity.

If we decide that the limiting percentage is to be 1 percent of e, then, the limiting slenderness ratio turns out to be

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If the actual slenderness ratio is greater than (l/k)2 , then use the secant formula

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EXAMPLE 10

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EXAMPLE 11

The figure shows a workpiece clamped to a milling machine table by a bolt tightened to a tension of 2000 lbf. The clamp contact is offset from the centroidal axis of the strut by a distance e = 0.10 in, as shown in part b of the gure. The strut, or block, is steel, 1 in square and 4 in long, as shown. Determine the maximum compressive stress in the block.

First we find A = bh = 1(1) = 1 in2, I = bh3/12 = 1(1)3/12 = 0.0833 in4, k2 = I/A = 0.0833/1 = 0.0833 in2, and l/k = 4/(0.0833)1/2 = 13.9. The limiting slenderness ratio is

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before it need be treated by using the secant formula.

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So the maximum compressive stress is

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