Strength of Materials by F L Singer 4 Ed Solutions

8/9/2019 Strength of Materials by F L Singer 4 Ed Solutions

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Design for Flexure and Shear

To determine the load capacity or the size of beam section, it must satisfy the allowable stresses in

both flexure (bending) and shear. Shearing stress usually governs in the design of short beams thatare heavily loaded, while flexure is usually the governing stress for long beams. In materialcomparison, timber is low in shear strength than that of steel.

For any cross-sectional shape, flexure and shear are given in the following formulas:

lexure ormula

!orizontal Shear Stress

For rectangular beam, the following defines for flexure and shear:

lexure formula for rectangular beam

!orizontal shear stress for rectangular beam

"heref b # flexure stressf v # bending stress

$ # maximum moment applied to the beam% # maximum shear applied to the beam

I # moment of inertia about the neutral axis

& # moment of areab # breadthd # depth

Solution to 'roblem * + esign for lexure and Shear

Problem 5!

- rectangular beam of width b and height h carries a central concentrated load ' on a simply


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supported span of length . /xpress the maximum f v in terms of maximum f b.

Solution 5!

From the figure:

From flexure formula:

From shear stress formula:

answer

Problem 5"

- laminated beam is composed of five plan0s, each 1 in. by 2 in., glued together to form a section 1in. wide by 3* in. high. The allowable shear stress in the glue is 4* psi, the allowable shear stress in

the wood is 32* psi, and the allowable flexural stress in the wood is 32** psi. etermine themaximum uniformly distributed load that can be carried by the beam on a 15ft simple span.

Solution 5"

Maximum moment for simple beam

Maximum shear for simple beam


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For bending stress of wood

For shear stress of wood

For shear stress in the glued joint

Where:

Thus,

Use wo = 1250 lb/ft for safe value of uniformly distributed load. answer

Problem 5#

ind the cross5sectional dimensions of the smallest s6uare beam that can be loaded as shown inig. '52 if f v 7 3.* $'a and f b 7 $'a.


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Solution 5#

ased on bending stress !s"uare b # d$:

ased on shear stress !s"uare b # d$:

Use 145 mm × 145 mm s"uare beam answer

Problem 5$

- wide5flange section having the dimensions shown in ig. '58 supports a distributed load of wo lb9ft on a simple span of length ft. etermine the ratio of the maximum flexural stress to the

maximum shear stress.


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Solution 5$

ending stress:

where:

Thus,

%hear stress:

where:

!see &omputation above$

Thus,

'atio !flexural stress : shear stress$

answer


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Problem 55

- simply supported beam of length carries a uniformly distributed load of 1*** :9m and has thecross section shown in ig. '5. ind to cause a maximum flexural stress of 31 $'a. "hat

maximum shearing stress is then developed;

Solution 55Flexural %tress

Wheref b # () M*a

M # (+ wo- # (+ !)///$- # 01/- 23m

& # 4!1/$ # (1 mm5 # 6//!1/6$+( 7 //!(1/6$+( # 668 601 /// mm8

Thus,

answer

%hearing %tress

Where

9 # 4 wo- # 4!)///$!0.11$ # )1/ 2

# (/ ///!(//$ ; !) 1/$!).1$ # ( 0( 1/ mm6

b # !1/$ # (// mm

Thus,


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answer

Problem 5%

The distributed load shown in ig. '51 is supported by a box beam having the same cross5sectionas that in 'rob. . etermine the maximum value of wo that will not exceed a flexural stress of 3*

$'a or a shearing stress of 3.* $'a.

Solution 5%

From shear diagram

ased on allowable bending stress

Where !From %olution 11$:& # (1 mm

5 # 668 601 /// mm8

Thus,

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ased on allowable shear stress

Where !From %olution 11$: # ( 0( 1/ mm6

5 # 668 601 /// mm8

b # (// mm

Thus,

For safe value of wo, use wo = 11.26 kN/m answer

Problem 5&

- beam carries two concentrated loads ' and triangular load of


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To draw the hear !"a#ram

(. 9 # 9= ; # * ; / # *9> # 9> 7 * # * 7 * # /

1. %hear at are re&tangular.

). %hear at = is paraboli& !nd degree

&urve$.

0. -o&ation of ?ero shear:y s"uared property of parabola

x + * # ( + 6*

x # ).@6 ft( 7 x # 1./0 ft

'o draw the (oment Diagram

3. $ - # *

2. $> # $ - ? -rea in shear diagram

$> # * 5 8' # 58' lb@ft

? -rea in shear diagram$/ # 58' ? 29< (1.4


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. $ # $A ? -rea in shear diagram$ # 58' ? 8' # *

1. The moment diagram at -> and A are straight lines (3st degree curves) while at >A is ased on allowable bending stress

"here (rom Solution ==)c # 1 in

I # ased on allowable shear stress

"here (rom Solution ==)& #


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Solution 5

From the shear diagram

Maximum moment # sum of area in %hear >iagram at the left of point of ?ero shear

ased on allowable flexural stress

Where& # (1/ mm5 # //!6//6$+( 7 (01!1/6$+(

5 # (61 8().)0 mm8

Thus,


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Where # //!1$!(60.1$ ; (1!1$!).1$ # (.1 mm6

5 # (61 8().)0 mm8

b # 1 mm

Thus,

For safe value of wo, use wo = $4.%6 N/m. answer

Problem 5+

- channel section carries a concentrated loads " and a total distributed load of 8" as shown in ig.'54. %erify that the :- is 2.3= in. above the bottom and that I:- # 12 in8. Dse these values to

determine the maximum value of " that will not exceed allowable stresses in tension of 1,*** psi, incompression of 3*,*** psi, or in shear of ,*** psi.

Solution 5+

Hide Based on allowable bending stress:

By symmetry

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!okay!$

y transfer formula for moment of inertia


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!okay!$

For & = '2( lb)ft

Top fiber in tension

ottom fiber in &ompression

For & = ( lb)ft

Top fiber in &ompression

ottom fiber in tension

Based on allowable shear stress:


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Where

Thus,

For safe value of W, use ( = 4045 lb answer

Problem 5+!

- box beam carries a distributed load of 2** lb9ft and a concentrated load ' as shown in ig. '54*.etermine the maximum value of ' if f b 7 32** psi and f v 7 3* psi.


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Solution 5+!

=he&A M= from the overhang segment

!okay!$

ased on allowable bending stress

Where

M # .1* ; (0// lb3ft

& # (+ # ) in5 # (/!(6$+( 7 !(/6$+( # 006.66 in8

Thus,


Where

9 # /.1* ; (()/ lb


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# (/!($!1.1$ ; B 1!($!.1$ C # / in6

b # in

Thus,

For safe value of *, use * = %4+0 lb answer

Spacing of Eivets or >olts in >uilt5Dp >eams"hen two or more thin layers of beams are fastened together with a bolt or a rivet so that they act asa unit to gain more strength, it is necessary to design the to size or spacing of these bolts or rivets so

that it can carry the shearing force acting between each adFacent layers.

Aonsider the beam shown in the figure.


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The shearing stress at the contact surface between the two plan0s is

The effective area covered by each bolt group has a length e6ual to the spacing of the bolts. Thetotal shearing force acting between the two surfaces must be e6ual to the total shearing force E

produced by the bolts.

then,

where E is the total shearing force to be resisted by the bolts and is e6ual to the allowable shearingstress G area G number of bolts in the group. E should be ta0en at the contact surface nearest the

neutral axis where the shearing stress is greatest. The spacing of bolts, e, is also called pitch.

Problem 5+#

- wide flange section is formed by bolting together three plan0s, each * mm by 2** mm, arrangedas shown in ig. '542. If each bolt can withstand a shearing force of 0:, determine the pitch if the

beam is loaded so as to cause a maximum shearing stress of 3.8 $'a.


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Solution 5+#

Where

Thus,

%pa&ing of bolts

answer

Problem 5+

- box beam, built up as shown in ig. '54


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the maximum value of ' that will not exceed f v # 32* psi in the beam or a shearing force of


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Where:

9 # +6 *

2


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>ased on allowable flexure stressH

>ased on shear stress of woodH

>ased on shear strength of screwsH


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or safe value of wo, use wo # 5"*+$ lbft. answer

Problem 5+5

- concentrated load ' is carried at midspan of a simply supported 325ft span. The beam is made of 25in. by 15in. pieces screwed together, as shown in ig. '54. If the maximum flexural stress

developed is 38** psi, find the maximum shearing stress and the pitch of the screws if each screwcan resist 2** lb.

Solution 5+5

For concentrated load P at midspan of a simply supported beam of span L = 12 ft.

From the &ross se&tion shown:


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From bending stress

Maximum shear stress

answer

From strength of s&rews

answer

Problem 5+%

Three plan0s 8 in by 1 in., arranged as shown in ig. '541 and secured by bolts spaced 3 ft apart,are used to support a concentrated load ' at the center of a simply supported span 32 ft long. If '

causes a maximum flexural stress of 32** psi, determine the bolt diameters, assuming that theshear between the plan0s is transmitted by friction only. The bolts are tightened to a tension of 2* 0si

and the coefficient of friction between the plan0s is *.8*.


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Solution 5+%

From allowable flexural stress

%trength of bolt


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2ormal for&e

From tensile stress of bolt:

answer

Problem 5+&

- plate and angle girder similar to that shown in ig. 5


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'ivet &apa&ity in terms of shear !double shear$

'ivet &apa&ity in terms of bearing !use D b # / M*a$

Use ' # )(.) A2 for safe value of '

From the strength of rivet

answer

Problem 5+

-s shown in ig. '54, two A


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Solution 5+

From Illustratie Problem !"1

# = 1$$ %Pa s&ear stress

'b = 22$ %Pa bearing stress for single s&ear riet

'b = 2($ %Pa bearing stress for double s&ear riet


'ivet &apa&ity in shear !single shear$

'ivet &apa&ity in bearing !use D b # / M*a$

Use ' # 1).0/1 A2 for safe value of '

From strength of rivets


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answer

Problem 5++ - beam is formed by bolting together two "2** G 3** sections as shown in ig. '544. It is used to

support a uniformly distributed load of


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Maximum moment

Maximum shear


Maximum flexural stress

answer

olt pit&h

answer

Ahapter *1 5 >eam eflections

Deflection of .eams

The deformation of a beam is usually expressed in terms of its deflection from its original unloadedposition. The deflection is measured from the original neutral surface of the beam to the neutral

surface of the deformed beam. The configuration assumed by the deformed neutral surface is 0nown


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as the elastic curve of the beam.

(ethods of Determining .eam Deflections

:umerous methods are available for the determination of beam deflections. These methods includeH

3. ouble5integration method

2. -rea5moment method

eam eflections

The double integration method is a powerful tool in solving deflection and slope of a beam at anypoint because we will be able to get the e6uation of the elastic curve.

In calculus, the radius of curvature of a curve y # f(x) is given by

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In the derivation of flexure formula, the radius of curvature of a beam is given as

eflection of beams is so small, such that the slope of the elastic curve dy9dx is very small, ands6uaring this expression the value becomes practically negligible, hence

Thus, /I 9 $ # 3 9 y

If /I is constant, the e6uation may be written asH

where x and y are the coordinates shown in the figure of the elastic curve of the beam under load, yis the deflection of the beam at any distance x. / is the modulus of elasticity of the beam, I representthe moment of inertia about the neutral axis, and $ represents the bending moment at a distance x

from the end of the beam. The product /I is called the flexural rigidity of the beam.

The first integration y yields the slope of the elastic curve and the second integration y gives thedeflection of the beam at any distance x. The resulting solution must contain two constants of

integration since /I yK # $ is of second order. These two constants must be evaluated from 0nownconditions concerning the slope deflection at certain points of the beam. or instance, in the case of

a simply supported beam with rigid supports, at x # * and x # , the deflection y # *, and in locatingthe point of maximum deflection, we simply set the slope of the elastic curve y to zero.

Problem %!5

etermine the maximum deflection E in a simply supported beam of length carrying a concentratedload ' at midspan.

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Solution %!5


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The negative sign indi&ates that the defle&tion is below the undeformed neutral axis.

Therefore,

answer

Problem %!%

etermine the maximum deflection E in a simply supported beam of length carrying a uniformlydistributed load of intensity wo applied over its entire length.

Solution %!%

From t&e gure below


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Therefore,

Maximum defle&tion will o&&ur at x # 4 - !midspan$

answer

TaAing W # wo-:

answer

Problem %!&

etermine the maximum value of /Iy for the cantilever beam loaded as shown in ig. '51*=. Ta0ethe origin at the wall.

Solution %!&

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that varies from zero at the wall to wo at the free end. Ta0e the origin at the wall.

Solution %!

y ratio and proportion


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answer

Problem %"!

The simply supported beam shown in ig. '513* carries a uniform load of intensity wo symmetricallydistributed over part of its length. etermine the maximum deflectionE and chec0 your result by

letting a # * and comparing with the answer to 'roblem 1*1.

Solution %"!

By symmetry

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!okay!$

Problem %""

Aompute the value of G5 E at midspan for the beam loaded as shown in ig. '5133. If / # 3* M'a,what value of 5 is re6uired to limit the midspan deflection to 39


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Solution %"#


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Therefore,


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Thus,

answer

Problem %"$

or the beam loaded as shown in ig. '5138, calculate the slope of the elastic curve over the rightsupport.

Solution %"$

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Problem %"5

Aompute the value of /I y at the right end of the overhanging beam shown in ig. '513.

Solution %"5

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answer

Defection under the load P: 4note / = a 5 b = L6

answer

Part /b0: $aximum deflection between the supports

The maximum deflection between the supports will occur at the point where y # *.

-t y # *, 〈 x 5 a 〉 do not exist thus,


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-t ,

answer

Problem %"&

Eeplace the load ' in 'rob. 131 by a cloc0wise couple $ applied at the right end and determine theslope and deflection at the right end.

Solution %"&

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answer

Problem %"

- simply supported beam carries a couple $ applied as shown in ig. '513. etermine thee6uation of the elastic curve and the deflection at the point of application of the couple. Then lettinga # and a # *, compare your solution of the elastic curve with cases 33 and 32 in theSummary of

>eam oadings.

Solution %"

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answer

When a # / !moment load is at the left support$:

answer

When a # - !moment load is at the right support$:


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answer

Problem %"+

etermine the value of /Iy midway between the supports for the beam loaded as shown in ig. '5

134.

Solution %"+


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By ratio and proportion:

y symmetry:


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Therefore,


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and the area and location of centroid are defined as follows.

1antilever 2oadings

- # area of moment diagram$x # moment about a section of distance x

barred x # location of centoidegree # degree power of the moment diagram

1ouple or (oment 2oad

egreeH zero


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1oncentrated 2oad

egreeH first

3niformly Distributed 2oad

egreeH second

3niformly 4arying 2oad

egreeH third

Problem %#$

or the beam loaded as shown in ig. '5128, compute the moment of area of the $ diagramsbetween the reactions about both the left and the right reaction.


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Solution %#$

$oment diagram by parts can be drawn in different waysL three are shown below.


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Problem %#5

or the beam loaded as shown in ig. '512, compute the moment of area of the $ diagramsbetween the reactions about both the left and the right reaction. (!intH raw the moment diagram by

parts from right to left.)


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Solution %#5


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answer

answer

Problem %#%

or the eam loaded as shown in ig. '5121, compute the moment of area of the $ diagramsbetween the reactions about both the left and the right reaction.


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Solution %#%

By symmetry

and

answer


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Thus,

answer

Problem %#&

or the beam loaded as shown in ig. '512=compute the moment of area of the $ diagramsbetween the reactions about both the left and the right reaction. (!intH Eesolve the trapezoidal

loading into a uniformly distributed load and a uniformly varying load.)

Solution %#&


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answer

answer

Problem %#

or the beam loaded with uniformly varying load and a couple as shown in ig. '512 compute themoment of area of the $ diagrams between the reactions about both the left and the right reaction.


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Solution %#


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answer

answer

Problem %#+Solve 'rob. 12 if the sense of the couple is counterclocwise instead of clocwise as shown in

ig. '512.

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Solution %#+


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answer

answer

Problem %!

or the beam loaded as shown in ig. '51)barred(N) - . rom theresult determine whether the tangent drawn to the elastic curve at > slopes up or down to the right.

(!intH Eefer to the deviation e6uations and rules of sign.)

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Solution %!

answer

The value of !


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Solution %"

Hide 7lic8 &ere to s&ow or &ide t&e solution

answer

The uniform load over span


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Problem %#

or the beam loaded as shown in ig. '51) barred(N) -. rom thisresult, is the tangent drawn to the elastic curve at > directed up or down to the right; (!intH Eefer to

the deviation e6uations and rules of sign.)

Solution %#

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answer

The value of !


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'heorems of 6rea-(oment (ethod

Theorem IThe change in slope between the tangents drawn to the elastic curve at any two points - and > is

e6ual to the product of 39/I multiplied by the area of the moment diagram between these two points.

Theorem IIThe deviation of any point > relative to the tangent drawn to the elastic curve at any other point -, ina direction perpendicular to the original position of the beam, is e6ual to the product of 39/I

multiplied by the moment of an area about > of that part of the moment diagram between points -and >.

and

7ules of Sign


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Solution %%

answer

Problem %&

or the beam loaded as shown in ig. '51


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Solution %&


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Thus, E # K t+= K # /.00@) in answer

Problem %

or the cantilever beam shown in ig. '51


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9 G5E # ).)0 A23m6 upward answer

Problem %+

The downward distributed load and an upward concentrated force act on the cantilever beam in ig.'51


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9 The free end will move by /./@@8 in&h downward. answer

Problem %$!

Aompute the value of E at the concentrated load in 'rob. 1


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Solution %$!Hide 7lic8 &ere to s&ow or &ide t&e solution

9 E # /./)@( in&h downward answer

Problem %$"

or the cantilever beam shown in ig. '5183, what will cause zero deflection at -;



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Solution %$"

answer

Problem %$#

ind the maximum deflection for the cantilever beam loaded as shown in igure '5182 if the crosssection is * mm wide by 3* mm high. Dse / # 14 M'a.


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Solution %$#




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9 Emax # mm answer

Problem %$ind the maximum value of /IE for the cantilever beam shown in ig. '518


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Therefore

answer

Problem %$$

etermine the maximum deflection for the beam loaded as shown in ig. '5188.

Solution %$$Hide 7lic8 &ere to s&ow or &ide t&e solution



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Therefore

answer

Problem %$5

Aompute the deflection and slope at a section < m from the wall for the beam shown in ig. '518. -ssume that / # 3* M'a and 5 #


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Solution %$5




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Therefore:

answer

answer

Problem %$%

or the beam shown in ig. '5181, determine the value of 5 that will limit the maximum deflection to

*.* in. -ssume that / # 3. G 3*

1

psi.


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Solution %$%




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Problem %$&

ind the maximum value of /IE for the beam shown in ig. '518=.

Solution %$&




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Therefore

answer

Problem %$

or the cantilever beam loaded as shown in ig. '518, determine the deflection at a distance x fromthe support.

Solution %$




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&omets about :

Triangular for&e to the left of :

Triangular upward for&e:

'e&tangle !wo by x$:

'ea&tions ' and M:

!e"at"o at w"th the ta#et l"e throu#h 3


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Therefore,

answer

5eflections in Simply Supported >eams + -rea5$oment $ethod

The deflection E at some point > of a simply supported beam can be obtained by the following stepsH

3. Aompute

2. Aompute


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Solution %5


By symmetry:

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From the figure

Thus

answer

Problem %5$

or the beam in ig. '518, find the value of /IE at 2 ft from E2. (!intH raw the reference tangent tothe elastic curve at E2.)

Solution %5$Hide 7lic8 &ere to s&ow or &ide t&e solution

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y ratio and proportion:


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answer

Problem %55

ind the value of /IE under each concentrated load of the beam shown in ig. '51.

Solution %55


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>efle&tions:


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L answer

answer

Problem %5%

ind the value of /IE at the point of application of the 2** :@m couple in ig. '511.

Solution %5%


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answer

Problem %5&

etermine the midspan value of /IE for the beam shown in ig. '51=.

Solution %5&Hide 7lic8 &ere to s&ow or &ide t&e solution



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answer

Problem %5

or the beam shown in ig. '51, find the value of /IE at the point of application of the couple.


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Solution %5





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The negative sign indi&ates that the defle&tion is opposite to the dire&tion sAet&hed in the figure.

Thus,

upward answer

Problem %5+

- simple beam supports a concentrated load placed anywhere on the span, as shown in ig. '514.$easuring x from -, show that the maximum deflection occurs at x # PB(2 5 b2)9


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Solution %5+




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From the figure:

!okay!$

Problem %%!

- simply supported beam is loaded by a couple $ at its right end, as shown in ig. '511*. Show thatthe maximum deflection occurs at x # *.==.


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Solution %%!




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From the figure

!okay!$

Problem %%"

Aompute the midspan deflection of the symmetrically loaded beam shown in ig. '5113. Ahec0 your answer by letting a # 92 and comparing with the answer to 'roblem 1*4.

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Solution %%"


answer

When a # 4-



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L answer

!okay!$

Problem %%#

etermine the maximum deflection of the beam shown in ig. '5112. Ahec0 your result by letting a #92 and comparing with case in Table 152. -lso, use your result to chec0 the answer to'rob. 1


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answer


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=he&A *roblem )16:

wo # )// 2+mJ - # 1 mJ a # m

!okay!$

When a # -+ !the load is over the entire span$

Therefore

Problem %%

etermine the maximum deflection of the beam carrying a uniformly distributed load over the middleportion, as shown in ig. '511


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answer

When b # -J b # 4-

!okay!$

Problem %%$

The middle half of the beam shown in ig. '5118 has a moment of inertia 3. times that of the rest of the beam. ind the midspan deflection. (!intH Aonvert the $ diagram into an $9/I diagram.)

Solution %%$




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Therefore,

answer

Problem %%5

Eeplace the concentrated load in 'rob. 118 by a uniformly distributed load of intensity wo acting over

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the middle half of the beam. ind the maximum deflection.

Solution %%5




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Therefore,

answer

Problem %%%

etermine the value of /IE at the right end of the overhanging beam shown in ig. '5111.

Solution %%%




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Solution %%&




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The negative sign indi&ates that the elasti& &urve is below the tangent line. 5t is shown in the

figure indi&ated as t=+. %ee 'ules of %ign for


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Solution %%




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answer

Thus,

Under the (//7lb load:

The negative sign indi&ates that the elasti& &urve is below the referen&e tangent.

Therefore,

downward answer

Problem %%+

Aompute the value of /IE midway between the supports of the beam shown in ig. '5114.

Solution %%+




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y s"uared property of parabola:


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With the values of G5 t=+


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Solution %&!




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The negative signs above indi&ates only the lo&ation of elasti& &urve relative to the referen&e

tangent. 5t does not indi&ate magnitude. 5t shows that the elasti& &urve is below the referen&e

tangent at points < and =.



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answer

$idspan eflection + eflections in Simply Supported >eamsIn simply supported beams, the tangent drawn to the elastic curve at the point of maximum

deflection is horizontal and parallel to the unloaded beam. It simply means that the deviation fromunsettling supports to the horizontal tangent is e6ual to the maximum deflection. If the simple beam

is symmetrically loaded, the maximum deflection will occur at the midspan.

inding the midspan deflection of a symmetrically loaded simple beam is straightforward because itsvalue is e6ual to the maximum deflection. In unsymmetrically loaded simple beam however, themidspan deflection is not e6ual to the maximum deflection. To deal with unsymmetrically loadedsimple beam, we will add a symmetrically placed load for each load actually acting on the beam,

ma0ing the beam symmetrically loaded. The effect of this transformation to symmetry will double theactual midspan deflection, ma0ing the actual midspan deflection e6ual to one5half of the midspan

deflection of the transformed symmetrically loaded beam.

Problem %&

or the beam shown in ig. '51=


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!okay!$


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Problem %&$

ind the deflection midway between the supports for the overhanging beam shown in ig. '51=8.

Solution %&$


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answer

Problem %&5

Eepeat 'rob. 1=8 for the overhanging beam shown in ig. '51=.

Solution %&5


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answer

Problem %&%etermine the midspan deflection of the simply supported beam loaded by the couple shown in ig.

'51=1.

Solution %&%


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answer

Problem %&&etermine the midspan deflection of the beam loaded as shown inig. '51==.

Solution %&&


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answer

Hide +not&er olution



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!okay!$

Problem %&etermine the midspan value of /IE for the beam shown in ig. '51=.

Solution %&


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answer

Problem %&+

etermine the midspan value of /IE for the beam shown in ig. '51=4 that carries a uniformlyvarying load over part of the span.

Solution %&+


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answer

Problem %!

etermine the midspan value of /IE for the beam loaded as shown in ig. '51*.

Solution %!

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answer

Problem %"

Show that the midspan value of /IE is (wob98)(


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Solution %"


Part (a)



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answer

*art b

answer

$ethod of Superposition + >eam eflection

The slope or deflection at any point on the beam is e6ual to the resultant of the slopes or deflectionsat that point caused by each of the load acting separately.


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7otation and Deflection for 1ommon 2oadings

1ase ": 1oncentrated load at the free end of cantilever

beam

$aximum $oment

Slope at end

$aximum deflection

eflection /6uation ( is positive downward)

1ase #: 1oncentrated load at any point on the span of

cantilever beam

$aximum $oment

Slope at end

$aximum deflection


1ase : 3niformly distributed load over the entire length of cantilever beam

$aximum $oment


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Slope at end

$aximum deflection


1ase $: 'riangular load, full at the fixed end and 8ero at

the free end, of cantilever beam

$aximum $oment

Slope at end

$aximum deflection


1ase 5: (oment load at the free end of cantilever beam

$aximum $oment

Slope at end

$aximum deflection



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1ase %: 1oncentrated load at the midspan of simple

beam

$aximum $oment

Slope at end

$aximum deflection


1ase &: 1oncentrated load at any point on simple beam

$aximum $oment

Slope at end

$aximum deflection

eflection at the center (not maximum)



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1ase : 3niformly distributed load over the entire span

of simple beam

$aximum $oment

Slope at end

$aximum deflection


1ase +: 'riangle load with 8ero at one support and full at

the other support of simple beam

$aximum $oment

Slope at end

$aximum deflection


1ase "!: 'riangular load with 8ero at each support and full at the midspan of simple beam

$aximum $oment


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Slope at end

$aximum deflection


1ase "": (oment load at the right support of simple

beam

$aximum $oment

Slope at end

$aximum deflection



1ase "#: (oment load at the left support of simple beam

$aximum $oment

Slope at end


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$aximum deflection



Problem %5

etermine the midspan value of /IE for the beam loaded as shown in ig. '51. Dse the method of superposition.

Solution %5


From 7ase ;o. , of ummary of Beam Loadings< de3ection at t&e center is

Thus, for Fig. *7)1

G5 Emidspan # G5 Emidspan due to (// lb for&e ; G5 Emidspan due to / lb for&e

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answer

Problem %%

etermine the value of /IE under each concentrated load in ig. '511.

Solution %%


From 7ase ;o. , of ummary of Beam Loadings< t&e de3ection euations are

The point under the load is generally lo&ated at and at this point, both e"uations above

will be&ome

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!eflet"o uder the 500 N load

G5E # G5E due to 1// 2 load ; G5E due to // 2 load

answer

!eflet"o uder the +00 N load

G5E # G5E due to 1// 2 load ; G5E due to // 2 load

answer

Problem %&

etermine the midspan deflection of the beam shown in ig. '51= if / # 3* M'a and I # 2* G3*1 mm8.

Solution %&


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From 7ase ;o. ,< midspan de3ection is

From =ase 2o. , midspan defle&tion is

&"ds-a deflet"o of the #"e beam

G5E # G5E due to A2 &on&entrated load ; G5E due to ( A2+m uniform loading

answer

Problem %

etermine the midspan value of /IE at the left end of the beam shown in ig. '51.

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Solution %

!ideAlic0 here to show or hide the solution

rom the figure below, the total deformation at the end of overhang is

The rotation N at the left support is combination ofAase :o. 32 and by integration of Aase :o. =.

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Solving for

/IN # /IN due to ** :@m moment at left support 5 /IN due to 8** :9m uniform load

-pply Aase :o.


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'otal deflection at the free end

answer

!ide-nother Solution (-rea5moment method)

This problem can be done with less effort by area5moment method.



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The negative sign above indicates that the elastic curve is below the tangent line.

(okay! )

Problem %+The beam shown in ig. '514 has a rectangular cross section 8 inches wide by inches deep.Aompute the value of ' that will limit the midspan deflection to *. inch. Dse / # 3. G 3*1 psi.

Solution %+

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>&e oer&ang is resoled into simple beam wit& end moments. >&e magnitude of

end moment is<

Moment of inertia of beam se&tion

The midspan defle&tion is a &ombination of defle&tion due to uniform load and two end

moments. Use =ase 2o. and =ases 2o. , ((, and ( to solve for the midspan defle&tion.



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Type of Loading Midspan Deection


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answer

Problem %+!

The beam shown in ig. '514* has a rectangular cross section * mm wide. etermine the proper depth d of the beam if the midspan deflection of the beam is not to exceed 2* mm and the flexural

stress is limited to 3* $'a. Dse / # 3* M'a.

Solution %+!


ased on allowable flexural stress



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ased on allowable midspan defle&tion. Use =ase 2o. 0, the midspan defle&tion of simple beamunder &on&entrated load is given by

For the given beam, the midspan defle&tion is the sum of the midspan defle&tion of ea&h load

a&ting separately.

Use d = 1%5.%6 mm answer

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Problem %+"

etermine the midspan deflection for the beam shown in ig. '5143. (!intH -pplyAase :o. = andintegrate.)

Solution %+"


From 7ase ;o. ,< t&e midspan de3ection is

For the given beam

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answer

Problem %+#

ind the value of /IE midway between the supports for the beam shown in ig. '5142. (!intHAombine Aase :o. 33 and one half of Aase :o. .)

Solution %+#


>&e midspan de3ection from 7ase ;o. ( and 7ase ;o. 11 are respectiely<



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The given beam is transformed into a simple beam with end moment at the right support due to

the load at the overhang as shown in the figure below.

G5E # 4 of G5E due to uniform load over the entire span 7 G5E due to end moment


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answer

Problem %+

etermine the value of /IE at the left end of the overhanging beam in ig. '514&e rotation at t&e left support is t&e combination of 7ase ;o. , and 7ase ;o. 12.



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The overhang beam is transformed into a simple beam and the end moment at the free end of the

overhang is &arried to the left support of the transformed beam.

The negative sign indi&ates that the rotation at the left end &ontributed by the end moment !taAen

as negative$ is greater than the rotation at the left end &ontributed by the &on&entrated load !taAen

as positive$.

From =ase 2o. 1, the end defle&tion is

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The defle&tion at the overhang due to moment load alone is

Total defle&tion at the left end of the given beam is

answer

Problem %+$

The frame shown in ig. '5148 is of constant cross section and is perfectly restrained at its lower end. Aompute the vertical deflection caused by the couple $.

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!ide Alic0 here to read or hide Solution 148

answer

Problem %+5

Solve 'roblem 148 if the couple is replaced by a downward load '.

!ide Alic0 here to read or hide Solution 14

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answer

Problem %+%

In ig. '5141, determine the value of ' for which the deflection under ' will be zero.

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Hide 7lic8 &ere to read or &ide olution 0"0

+pply 7ase ;o. ( and 7ase ;o. 11 to nd t&e slope at t&e rig&t support.

Use =ase 2o. ( for the defle&tion at the free end due to &on&entrated load *.

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answer

Problem %+&

or the beam in 'rob. 141, find the value of ' for which the slope over the right support will be zero.

From olution 0"0<

answer

AonFugate >eam $ethod + >eam eflectionSlope on real beam # Shear on conFugate beam

eflection on real beam # $oment on conFugate beam


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Properties of 1on9ugate .eam

/ngr. Ahristian Jtto $ohr

3. The length of a conFugate beam is always e6ual to the length of the actual beam.

2. The load on the conFugate beam is the $9/I diagram of the loads on the actual beam.


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xamples of .eam and its 1on9ugate

The following are some examples of beams and its conFugate. oadings are omitted.


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Problem %5

Aompute the midspan value of /IE for the beam shown in ig. '51


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Solution %5 /3sing (oment Diagram by Parts0


By symmetry<

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The loads of &onjugate beam are symmetri&al, thus,

For this beam, the maximum defle&tion will o&&ur at the midspan.


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Therefore, the maximum defle&tion is

below the neutral axis answer

6nother Solution /3sing the 6ctual (oment Diagram0


4Conjugate beam method using the actual moment diagram6



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y symmetry of &onjugate beam


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The maximum defle&tion will o&&ur at the midspan of this beam

Therefore, the maximum defle&tion is

below the neutral axis okay!

Problem %5$

or the beam in ig. '518, find the value of /IE at 2 ft from E2.

Solution %5$


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oling for reactions

From the &onjugate beam



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Thus, the defle&tion at is

downward answer

Problem %55

ind the value of /IE under each concentrated load of the beam shown in ig. '51.


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Solution %55




190/314




191/314

=onsider the se&tion to the left of in &onjugate beam

Thus, the defle&tion at is

answer

=onsider the se&tion to the right of = in &onjugate beam


192/314

Thus, the defle&tion at = is

downward answer

Problem %5%

ind the value of /IE at the point of application of the 2** :@m couple in ig. '511.

Solution %5%


From t&e real beam



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194/314

Therefore, the defle&tion at = is

downward answer

Problem %5&

etermine the midspan value of /IE for the beam shown in ig. '51=.

Solution %5&


From t&e load diagram



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From the moment diagram



196/314

Thus, the defle&tion at the midspan is

below the neutral axis answer

Problem %5

or the beam shown in ig. '51, find the value of G5E at the point of application of the couple.

Solution %5




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198/314

Thus,

answer

Strain /nergy $ethod (AastiglianoRs Theorem) + >eam eflectionItalian engineer -lberto Aastigliano (38= 38) developed a method of determining deflection of

structures by strain energy method. !is Theorem of the Derivatives of Internal Work of Deformation extended its application to the calculation of relative rotations and displacements

between points in the structure and to the study of beams in flexure.

Energy of structure is its capacity of doing wor0 and strain energy is the internal energy in the

structure because of its deformation. >y the principle of conservation of energy,


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where denotes the strain energy and represents the wor0 done by internal forces. T

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Strength of Materials by F L Singer 4 Ed Solutions

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