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Chapter 4

Deflection and Stiffness

Lecture Slides

The McGraw-Hill Companies 2012

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Chapter Outline

Shigleys Mechanical Engineering Design

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Force vs Deflection

Elasticityproperty of a material that enables it to regain its

original configuration after deformation Springa mechanical element that exerts a force when

deformed


Linear spring

Nonlinear

stiffening spring

Nonlinear

softening spring

Fig. 41

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Spring Rate

Relation between force and deflection,F=F(y)

Spring rate

For linear springs, k is constant, calledspring constant


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Axially-Loaded Stiffness

Total extension or contraction of a uniform bar in tension or

compression

Spring constant, with k=F/d


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Torsionally-Loaded Stiffness

Angular deflection (in radians) of a uniform solid or hollow

round bar subjected to a twisting moment T

Converting to degrees, and includingJ= pd4/32 for round solid

Torsional spring constant for round bar


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Deflection Due to Bending

Curvature of beam subjected to bending momentM

From mathematics, curvature of plane curve

Slope of beam at any pointx along the length

If the slope is very small, the denominator of Eq. (4-9)

approaches unity.

Combining Eqs. (4-8) and (4-9), for beams with small slopes,


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Recall Eqs. (3-3) and (3-4)

Successively differentiating


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(4-10)

(4-11)

(4-12)

(4-13)

(4-14)

Fig. 42

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Example 4-1


Fig. 42

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Example 4-1


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Example 4-1


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Example 4-1


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Beam Deflection Methods

Some of the more common methods for solving the integration

problem for beam deflection

Superposition

Moment-area method

Singularity functions

Numerical integration Other methods that use alternate approaches

Castigliano energy method

Finite element software


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Beam Deflection by Superposition

Superposition determines the effects of each load separately,then adds the results.

Separate parts are solved using any method for simple loadcases.

Many load cases and boundary conditions are solved andavailable in Table A-9, or in references such asRoarks

Formulas for Stress and Strain. Conditions

Each effect is linearly related to the load that produces it.

A load does not create a condition that affects the result of

another load. The deformations resulting from any specific load are not

large enough to appreciably alter the geometric relations ofthe parts of the structural system.


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Example 4-2


Fig. 43

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Example 4-2


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Example 4-2


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Example 4-3


Fig. 44

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Example 4-3


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Example 4-3


Fig. 44

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Example 4-3


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Example 4-4


Fig. 45

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Example 4-4


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Example 4-4


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Beam Deflection by Singularity Functions

A notation useful

for integrating

across

discontinuities

Angle brackets

indicate special

function todetermine whether

forces and moments

are active


Table 31

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Example 4-5


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Example 4-5


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Example 4-5


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Example 4-5


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Example 4-6


Fig. 46

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Example 4-6


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Example 4-6


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Example 4-6


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Example 4-7


Fig. 47

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Example 4-7

Shigleys Mechanical Engineering DesignFig. 47

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Example 4-7


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Example 4-7


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Example 4-7


4

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Example 4-7


E l 4 7

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Example 4-7


E l 4 7

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Example 4-7


St i E

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Strain Energy

External work done on elastic member in deforming it is

transformed intostrain energy, orpotential energy.

Strain energy equals product of average force and deflection.


S C St i E F l

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Some Common Strain Energy Formulas

For axial loading, applying k=AE/lfrom Eq. (4-4),

For torsional loading, applying k= GJ/lfrom Eq. (4-7),


S C St i E F l

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For direct shear loading,

For bending loading,


S C St i E F l

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For transverse shear loading,

where Cis a modifier dependent on the cross sectional shape.


S mmar of Common Strain Energ Form las

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Summary of Common Strain Energy Formulas


Example 4 8

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Example 4-8


Fig. 49

Example 4 8

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Example 4-8


Fig. 49

Example 4 8

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Example 4-8


Castiglianos Theorem

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Castiglianos Theorem

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.

For rotational displacement, in radians,


Example 4 9

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Example 4-9


Fig. 49

Example 4 9

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Example 4-9


Utilizing a Fictitious Force

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Utilizing a Fictitious Force

Castiglianos method can be used to find a deflection at a point

even if there is no force applied at that point.

Apply a fictitious force Q at the point, and in the direction, of

the desired deflection.

Set up the equation for total strain energy including the energy

due to Q.

Take the derivative of the total strain energy with respect to Q.

Once the derivative is taken, Q is no longer needed and can be

set to zero.


Finding Deflection Without Finding Energy

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Finding Deflection Without Finding Energy

For cases requiring integration of strain energy equations, it is

more efficient to obtain the deflection directly without explicitly

finding the strain energy.

The partial derivative is moved inside the integral.

For example, for bending,

Derivative can be taken before integration, simplifying the math. Especially helpful with fictitious force Q, since it can be set to

zero after the derivative is taken.


Common Deflection Equations

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Common Deflection Equations


Example 4-10

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Example 4-10


Fig. 410

Example 4-10

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Example 4 10


Example 4-10

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Example 4 10


Example 4-10

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Example 4 10


Example 4-11

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Example 4 11


Fig. 411

Example 4-11

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Example 4 11


Example 4-11

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Example 4 11


Fig. 411

Example 4-11

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p


Example 4-11

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p


Example 4-11

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p


Example 4-11

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p


Deflection of Curved Members

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Consider case of thick curved member in bending (See Sec. 3-18)

Four strain energy terms due to

Bending momentM

Axial forceFq

Bending moment due toFq

(since neutral axis and centroidal

axis do not coincide)

Transverse shearFr



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Strain energy due to bending momentM


where rn is the radius of the neutral axis

Fig. 412


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Strain energy due to axial forceFq



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Strain energy due to bending moment due toFq



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Strain energy due to transverse shearFr



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Combining four energy terms

Deflection by Castiglianos method

General for any thick circular curved member, with appropriate

limits of integration



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For specific example in figure,



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Substituting and factoring,


Deflection of Thin Curved Members

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Forthin curved members, sayR/h > 10, eccentricity is small

Strain energies can be approximated with regular energy equations

with substitution ofRdqfordx AsR increases, bending component dominates all other terms


Example 4-12

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Fig. 413

Example 4-12

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Fig. 413

Example 4-12

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Example 4-12

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Example 4-12

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Example 4-12

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Example 4-13

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Example 4-13

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Example 4-13

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Example 4-13

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Example 4-13

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Shigleys Mechanical Engineering DesignFig. 414 (b)

Statically Indeterminate Problems

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A system is overconstrainedwhen it has more unknown support

(reaction) forces and/or moments than static equilibrium

equations.

Such a system is said to bestatically indeterminate.

The extra constraint supports are call redundant supports.

To solve, a deflection equation is required for each redundant

support.


Statically Indeterminate Problems

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Example of nested springs

One equation of static

equilibrium

Deformation equation

Use spring constant relation to

put deflection equation in terms

of force

Substituting into equilibrium

equation,


Fig. 415

Procedure 1 for Statically Indeterminate Problems

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1. Choose the redundant reaction(s)

2. Write the equations of static equilibrium for the remaining

reactions in terms of the applied loads and the redundant

reaction(s).

3. Write the deflection equation(s) for the point(s) at the locations

of the redundant reaction(s) in terms of the applied loads and

redundant reaction(s).

4. Solve equilibrium equations and deflection equations

simultaneously to determine the reactions.


Example 4-14

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Fig. 416

Example 4-14

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Example 4-14

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Example 4-14

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Example 4-14

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Example 4-14

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Example 4-14

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Procedure 2 for Statically Indeterminate Problems

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1. Write the equations of static equilibrium in terms of the applied

loads and unknown restraint reactions.

2. Write the deflection equation in terms of the applied loads andunknown restraint reactions.

3. Apply boundary conditions to the deflection equation consistent

with the restraints.

4. Solve the set of equations.


Example 4-15

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Example 4-15

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Example 4-15

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Example 4-15

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Example 4-15

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Example 4-15

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Example 4-15

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Compression Members

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ColumnA member loaded in compression such that either its

length or eccentric loading causes it to experience more than

pure compression Four categories of columns

Long columns with central loading

Intermediate-length columns with central loading

Columns with eccentric loading

Struts or short columns with eccentric loading


Long Columns with Central Loading

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WhenPreaches critical

load, column becomes

unstable and bendingdevelops rapidly

Critical load depends on end

conditions


Fig. 418

Euler Column Formula

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For pin-ended column,

critical load is given by

Euler column formula,

Applies to other end

conditions withaddition of constant C

for each end condition


(4-42)

(4-43) Fig. 418

Recommended Values for End Condition Constant

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Fixed ends are practically

difficult to achieve

More conservative values ofCare often used, as in Table 4-2


Table 42


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UsingI = Ak2, whereA is the area and kis the radius of

gyration, Euler column formula can be expressed as

l/kis theslenderness ratio, used to classify columns according

to length categories.

Pcr/A is the critical unit load, the load per unit area necessary to

place the column in a condition ofunstable equilibrium.


Euler Curve

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PlottingPcr/A vs l/k, with C = 1 gives curvePQR



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Tests show vulnerability to failure near point Q

Since buckling is sudden and catastrophic, a conservative

approach nearQ is desired

Point Tis usually defined such thatPcr/A = Sy/2, giving


Condition for Use of Euler Equation

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For (l/k) > (l/k)1, use Euler equation

For (l/k) (l/k)1, use a parabolic curve between Sy and T


Fig. 419

Intermediate-Length Columns with Central Loading

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For intermediate-length columns,

where (l/k) (l/k)1, use a parabolic

curve between Sy and T General form of parabola

If parabola starts at Sy, then a = Sy

If parabola fits tangent to Euler

curve at T, then

Results inparabolic formula, also

known asJ.B. Johnson formula


Fig. 419

Columns with Eccentric Loading

i ll l d d l

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For eccentrically loaded column

with eccentricity e,

M = -P(e+y)

Substituting into d2y/dx2=M/EI,

Solving with boundary conditions

y = 0 atx = 0 and atx = l


Fig. 420


A id h l/2

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At midspan wherex = l/2



Th i i i l d

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The maximum compressive stress includes

axial and bending

SubstitutingMmax from Eq. (4-48)

Using Syc as the maximum value ofsc, and

solving forP/A, we obtain thesecant column

formula


Secant Column Formula


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ec/k2 is the eccentricity ratio

Design charts of secant column formula for various eccentricity

ratio can be prepared for a given material strength


Fig. 421

Example 4-16

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Example 4-16

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Example 4-17

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Example 4-17

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Example 4-18

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Example 4-19

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Example 4-19

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Struts or Short Compression Members

Strut short member loaded in compression

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Strut - short member loaded in compression

If eccentricity exists, maximum stress is atB

with axial compression and bending.

Note that it is not a function of length

Differs from secant equation in that it assumes

small effect of bending deflection

If bending deflection is limited to 1 percent ofe,

then from Eq. (4-44), the limiting slenderness

ratio for strut is


Fig. 422

Example 4-20

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Example 4-20

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Elastic Stability

Be alert for buckling instability in structural members that are

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Be alert for buckling instability in structural members that are

Loaded in compression (directly or indirectly)

Long or thin

Unbraced

Instability may be

Local or global

Elastic or plastic

Bending or torsion


Fig. 424

Fig. 425

Shock and Impact

Impact collision of two masses with initial relative velocity

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Impactcollision of two masses with initial relative velocity

Shocka suddenly applied force or disturbance


Shock and Impact

Example of automobile collision

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Example of automobile collision

m1 is mass of engine

m2 is mass of rest of vehicle

Springs represent stiffnesses of various structural elements

Equations of motion, assuming linear springs

Equations can be solved for any impact velocity


Suddenly Applied Loading

Weight falls distance h and

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Weight falls distance h and

suddenly applies a load to a

cantilever beam Find deflection and force

applied to the beam due to

impact


Fig. 427 (a)


Abstract model considering

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Abstract model considering

beam as simple spring

From Table A-9, beam 1,k= F/y =3EI/l3

Assume the beam to be

massless, so no momentum

transfer, just energy transfer Loss of potential energy from

change of elevation is

W(h + d)

Increase in potential energyfrom compressing spring is

kd2/2

Conservation of energy

W(h +d) = k

d2

/2 Shigleys Mechanical Engineering Design

Fig. 427 (b)


Rearranging

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Rearranging

Solving ford

Maximum deflection

Maximum force

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