Files 2-Handouts Ch07

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2005 Pearson Education South Asia Pte Ltd

7. Transverse Shear

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CHAPTER OBJECTIVES

Develop a method for findingthe shear stress in a beamhaving a prismatic x-sectionand made from homogeneousmaterial that behaves in alinear-elastic manner

This method of analysis islimited to special cases of x-sectional geometry

Discuss the concept of shear flow, with shear stressfor beams and thin-walled members

Discuss the shear center


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7. Transverse Shear

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CHAPTER OUTLINE

1. Shear in Straight Members

2. The Shear Formula

3. Shear Stresses in Beams4. Shear Flow in Built-up Members

5. Shear Flow in Thin-Walled Members

6. *Shear Center


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7.1 SHEAR IN STRAIGHT MEMBERS

Shear V is the result of a transverse shear-stressdistribution that acts over the beams x-section.

Due to complementary property of shear,associated longitudinal shear stresses also actalong longitudinal planes of beam


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As shown below, if top and bottom surfaces of eachboard are smooth and not bonded together, thenapplication of load P will cause the boards to sliderelative to one another.

However, if boards are bonded together,

longitudinal shear stresses will develop and distortx-section in a complex manner


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As shown, whenshear V is applied,the non-uniformshear-straindistribution over x-

section will cause itto warp, i.e., notremain in plane.


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Recall that the flexure formula assumes that x-sections must remain planeand perpendicular tolongitudinal axis of beam after deformation

This is violatedwhen beam is subjected to bothbending and shear, we assume that the warping is

so small it can be neglected. This is true for aslender beam(small depth compared with itslength)

For transverse shear, shear-strain distributionthroughout the depth of a beam cannot be easilyexpressed mathematically



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Thus, we need to develop the formula for shearstress is indirectlyusing the flexure formula andrelationship between moment and shear(V = dM/dx)



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7.2 THE SHEAR FORMULA

By first principles, flexure formula and V = dM/dx,we obtain

=

VQ

It Equation 7-3

=shear stress in member at the

pt located a distanceyfromthe neutral axis. Assumed tobe constant and therefore

averagedacross the width tofmember

V =internal resultant shear force, determined frommethod of sections and equations of equilibrium



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By first principles, flexure formula and V = dM/dx,we get:

=

VQ

It Equation 7-3

I =moment of inertia of entire x-sectional area

computed about the neutral axist =width of the members x-sectional area,

measured at the pt where is to be determined

Q =Ay dA=yA, where Ais the top (or bottom)portion of members x-sectional area, defined from

section where tis measured, andyis distance ofcentroid ofA, measured from neutral axis



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The equation derived is called the shear formula

Since Eqn 7-3 is derived indirectly from the flexure

formula, the material must behave in a linear-elastic manner and have a modulus of elasticitythat is the samein tension and in compression

Shear stress in composite members can also beobtained using the shear formula

To do so, compute Qand Ifrom the transformedsectionof the member as discussed in section 6.6.Thickness tin formula remains the actual width tof

x-section at the pt where

is to be calculated



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7.3 SHEAR STRESSES IN BEAMS

Rectangular x-section

Consider beam to have rectangular

x-section of width band height hasshown.

Distribution of shear stress

throughout x-section can bedetermined by computing shearstress at arbitrary heightyfrom

neutral axis, and plotting thefunction. Hence,

Q= y2 b( )1

2

h2

4



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After deriving Qand applying the

shear formula, we have

= y2( )6V

bh3

h2

4

Equation 7-4 Eqn 7-4 indicates that

shear-stress distribution

over x-section is parabolic.



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Aty= 0, we have

max = 1.5VA Equation 7-5

By comparison, max is 50%greater than the average

shear stress determinedfrom avg = V/A.



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Wide-flange beam

A wide-flange beam consists

of two (wide) flanges and aweb.

Using analysis similar to a

rectangular x-section, theshear stress distributionacting over x-section is shown



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Wide-flange beam

The shear-stress distribution also varies

parabolically over beams depth Note there is a jump in shear stress at the flange-

web junction since x-sectional thickness changes

at this ptThe web carries significantly

more shear force than the flanges



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Limitations on use of shear formula One major assumption in the development of the

shear formula is that shear stress is uniformly

distributed over width tat section where shearstress is to be determined

By comparison with exact mathematical analysisbased on theory of elasticity, the magnitudedifference can reach 40%

This is especially so for the flange of a wide-flangebeam

S



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Limitations on use of shear formula

The shear formula will also give inaccurate results

for the shear stress at the flange-web junction of awide-flange beam, since this is a pt of sudden x-sectional change (stress concentration occurs

here) Furthermore, inner regions of flanges are free

boundaries, thus shear stress at these boundaries

should be zero

However, shear formula calculated at these pts will

not be zero

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Limitations on use of shear formula

Fortunately, engineers are often interested in the

average maximum shear stress, which occurs atthe neutral axis, where b/hratio is very small

Also, shear formula does not give accurate results

when applied to members having x-sections thatare short or flat, or at pts where the x-sectionsuddenly changes

It should also not be applied across a section thatintersects the boundary of a member at an angle

other than 90

o

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IMPORTANT

Shear forces in beams cause non-linear shear-strain distributions over the x-section, causing it towarp

Due to complementary property of shear stress,

the shear stress developed in a beam acts on boththe x-section and on longitudinal planes

The shear formula was derived by considering

horizontal force equilibrium of longitudinal shearstress and bending-stress distributions acting on aportion of a differential segment of the beam

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IMPORTANT

The shear formula is to be used on straightprismatic members made of homogeneousmaterial that has linear-elastic behavior. Also,internal resultant shear force must be directed

along an axis of symmetry for x-sectional area For beam having rectangular x-section, shear

stress varies parabolically with depth.

For beam having rectangular x-section, maximumshear stress is along neutral axis

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IMPORTANT

Shear formula should not be used to determineshear stress on x-sections that are short or flat, orat pts of sudden x-sectional changes, or at a pt onan inclined boundary

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Procedure for analysisInternal shear

Section member perpendicular to its axis at the ptwhere shear stress is to be determined

Obtain internal shear V at the section

Section properties

Determine location of neutral axis, and determine

the moment of inertia Iof entire x-sectional areaabout the neutral axis

Pass an imaginary horizontal section through the

pt where the shear stress is to be determined

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Procedure for analysisSection properties

Measure the width tof the area at this section Portion of area lying either above or below this

section is A.

Determine Qeither by integration, Q =Ay dA, orby using Q = yA.

Here,yis the distance of centroid ofA, measuredfrom the neutral axis. (TIP: Ais the portion of themembers x-sectional area being held onto the

member by the longitudinal shear stresses.)

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Procedure for analysisShear stress

Using consistent set of units, substitute data intothe shear formula and compute shear stress

Suggest that proper direction of transverse shear

stress be established on a volume element ofmaterial located at the pt where it is computed

acts on the x-section in the same direction as V.From this, corresponding shear stresses acting onthe three other planes of element can be

established

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EXAMPLE 7.3

Beam shown is made from two boards. Determinethe maximum shear stress in the glue necessary tohold the boards together along the seams wherethey are joined. Supports at Band Cexert onlyvertical reactions on the beam.

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EXAMPLE 7.3 (SOLN)

Internal shear

Support reactions and shear diagram for beam are

shown below. Maximum shear in the beam is 19.5 kN.

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EXAMPLE 7.3 (SOLN)

Section properties

The centroid and therefore the neutral axis will be

determined from the reference axis placed at bottomof the x-sectional area. Working in units of meters,we have

y= = ... = 0.120 myA

A

Thus, the moment of inertia, computed about the

neutral axis is,I= ... = 27.0(10-6) m4

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EXAMPLE 7.3 (SOLN)

Section properties

The top board (flange) is being held onto the bottom

board (web) by the glue, which is applied over thethickness t= 0.03m. Consequently Ais defined asthe area of the top board, we have

Q = yA= [(0.180 m 0.015 m 0.120 m]

(0.03 m)(0.150 m)

Q= 0.2025(10-3

) m3

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EXAMPLE 7.3 (SOLN)

Shear stress

Using above data, and applying shear formula

yieldsmax = = ... = 4.88 MPa

VQ

It

Shear stress acting at top ofbottom board is shown here.

It is the glues resistance to thislateral or horizontal shear stressthat is necessary to hold the

boards from slipping at support C.

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7.4 SHEAR FLOW IN BUILT-UP MEMBERS

Occasionally, in engineeringpractice, members are built-up fromseveral composite parts in order toachieve a greater resistance toloads, some examples are shown.

If loads cause members to bend,fasteners may be needed to keepcomponent parts from sliding relative

to one another.To design the fasteners, we need to

know the shear force resisted by

fastener along members length

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This loading, measured as a force per unit length,is referred to as the shear flowq.

Magnitude of shear flow along any longitudinalsection of a beam can be obtained using similardevelopment method for finding the shear stress in

the beam

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Thus shear flow isq = VQ/I Equation 7-6

q =shear flow, measured as a force per unit lengthalong the beam

V =internal resultant shear force, determined from

method of sections and equations of equilibriumI =moment of inertia of entire x-sectional area

computed about the neutral axis

Q =Ay dA=yA, where Ais the x-sectional areaof segment connected to beam at juncture where

shear flow is to be calculated, andyis distancefrom neutral axis to centroid ofA

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Note that the fasteners in (a) and (b) supports thecalculated value ofq

And in (c) each fastener supports q/2 In (d) each fastener supports q/3

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IMPORTANT Shear flow is a measure of force per unit length

along a longitudinal axis of a beam.

This value is found from the shear formula and isused to determine the shear force developed in

fasteners and glue that holds the varioussegments of a beam together

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EXAMPLE 7.4

Beam below is constructed from 4 boards gluedtogether. It is subjected to a shear ofV= 850 kN.Determine the shear flow at Band Cthat must beresisted by the glue.

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EXAMPLE 7.4 (SOLN)

Section properties

Neutral axis (centroid) is located from bottom of the

beam. Working in units of meters, we have

y=y A

y= ... = 0.1968 m

Moment of inertia aboutneutral axis is

I= ... = 87.52(10-6) m4

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EXAMPLE 7.4 (SOLN)

Section properties

Since the glue at Band Bholds the top board to

the beam, we have

Likewise, glue at Cand Choldsinner board to beam, so

QB=yBAB= [0.305 m 0.1968 m](0.250 m)(0.01 m)

QB

= 0.270(10-3) m3

QC=yCAC= ... = 0.01025(10-3) m3

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EXAMPLE 7.4 (SOLN)

Shear flow

For Band B, we have

Similarly, for Cand C,

qB= VQB/I= [850 kN(0.270(10-3) m3]/87.52(10-6) m4

qB= 2.62 MN/m

qC= VQC/I= [850 kN(0.0125(10-3) m3]/87.52(10-6) m4

qC= 0.0995 MN/m

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EXAMPLE 7.4 (SOLN)

qB= 1.31 MN/m

qC= 0.0498 MN/m

Shear flow

Since two seams are used to secure each board,

the glue per meter length of beam at each seammust be strong enough to resist one-half of eachcalculated value ofq.Thus

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7.5 SHEAR FLOW IN THIN-WALLED MEMBERS

We can use shear-flow equation q = VQ/Ito find theshear-flow distribution throughout a members x-sectional area.

We assume that the member has thin walls, i.e.,wall thickness is small compared with height or

width of member

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Because flange wall is thin, shear stress will notvary much over the thickness of section, and weassume it to be constant. Hence,

q = t Equation 7-7

We will neglect the verticaltransverse component of shearflow because it is approx. zerothroughout thickness of element

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To determine distribution of shear flow along topright flange of beam, shear flow is

Equation 7-8q = (b/2 x)Vt d

2I

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Similarly, for the web of the beam, shear flow is

Equation 7-9q = +0.5(d2

/4 y2

)

Vt

I

db

2[ ]

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Value ofqchanges over the x-section, since Qwillbe different for each area segment A

qwill vary linearly along segments (flanges) thatare perpendicular to direction ofV, andparabolically along segments (web) that are

inclined or parallel to V qwill always act parallel to the walls of the

member, since section on which qis calculated is

taken perpendicular to the walls

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Directional sense ofqis such thatshear appears to flow through the x-section, inward at beams top flange,

combining and then flowingdownward through the web, and thenseparating and flowing outward atthe bottom flange

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IMPORTANT If a member is made from segments having thin

walls, only the shear flow parallel to the walls ofmember is important

Shear flow varies linearly along segments that are

perpendicular to direction of shear V Shear flow varies parabolically along segments that

are inclined or parallel to direction of shear V

On x-section, shear flows along segments so thatit contributes to shear V yet satisfies horizontal andvertical force equilibrium

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EXAMPLE 7.7

Thin-walled box beam shown is subjected to shearof10 kN. Determine the variation of shear flowthroughout the x-section.

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EXAMPLE 7.7 (SOLN)

By symmetry, neutral axis passes through center ofx-section. Thus moment of inertia is

I= 1/12(6 cm)(8 cm)3

1/12(4 cm)(6 cm)3

= 184 cm4

Only shear flows at pts B, Cand Dneeds to be determined. For pt B,area A 0 since it can be thought oflocated entirely at pt B. Alternatively,

Acan also represent the entire x-sectional area, in which caseQB=yA= 0 sincey= 0.

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EXAMPLE 7.7 (SOLN)

For pt C, area Ais shown dark-

shaded. Here mean dimensions areused since pt Cis on centerline ofeach segment. We have

Because QB= 0, then qB= 0

QC=yA= (3.5 cm)(5 cm)(1 cm) = 17.5 cm3

qC

= VQC/I= ... = 95.1 N/mm

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EXAMPLE 7.7 (SOLN)

Shear flow at Dis computed using thethree dark-shaded rectangles shown.We have

QD=yA= ... = 30 cm3

qC= VQD/I= ... = 163 N/mm

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EXAMPLE 7.7 (SOLN)

Using these results, and symmetry of x-section,shear-flow distribution is plotted as shown.Distribution is linear along horizontal segments

(perpendicular to V) and parabolic along verticalsegments (parallel to V)

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*7.6 SHEAR CENTER

Previously, we assumed that internal shear Vwas applied along a principal centroidal axis ofinertia that alsorepresents the axis of symmetry

for the x-section

Here, we investigate the effect of applying the

shear along a principal centroidal axis that is notan axis of symmetry

When a force P is applied to a channel section

along the once vertical unsymmetrical axis thatpasses through the centroidCof the x-sectionalarea, the channel bends downwards and also

twist clockwise

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*7.6 SHEAR CENTER

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*7.6 SHEAR CENTER

When the shear-flow distribution is integrated overthe flange and web areas, a resultant force ofFfineach flange and a force ofV=Pin the web is

created If we sum the moments of these forces about pt A,

the couple (or torque) created by the flange forces

causes the member to twistTo prevent the twisting, we need to apply P at a pt

Olocated a distance efrom the web of the channel,

thusMA = Ffd = Pe e =(Ffd)/P


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*7.6 SHEAR CENTER

IMPORTANT Shear center is the pt through which a force can be

applied which will cause a beam to bend and yetnot twist

Shear center will always lie on an axis of symmetry

of the x-section Location of the shear center is only a function of

the geometry of the x-section and does not depend

upon the applied loading

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*7.6 SHEAR CENTER

Procedure for analysis

Shear-flow resultants

Magnitudes of force resultants that create amoment about pt A must be calculated

For each segment, determine the shear flow qat

an arbitrary pt on segment and then integrate qalong the segments length

Note that V will create a linear variation of shearflow in segments that are perpendicular to V and aparabolic variation of shear flow in segments that

are parallel or inclined to V

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*7.6 SHEAR CENTER


Shear-flow resultants

Determine the direction of shear flow through thevarious segments of the x-section

Sketch the force resultants on each segment of

the x-section Since shear center determined by taking the

moments of these force resultants about a pt (A),choose this pt at a location that eliminates themoments of as many as force resultants as

possible

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*7.6 SHEAR CENTER


Shear center

Sum the moments of the shear-flow resultantsabout pt A and set this moment equal to momentofV about pt A

Solve this equation and determine the moment-arm distance e, which locates the line of action ofV from pt A

If axis of symmetry for x-section exists, shearcenter lies at the pt where this axis intersects line

of action ofV

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*7.6 SHEAR CENTER


Shear center

If no axes of symmetry exists, rotate the x-sectionby 90o

and repeat the process to obtain anotherline of action for V

Shear center then lies at the pt of intersection of thetwo 90

olines

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EXAMPLE 7.8

Determine the location of the shear center for thethin-walled channel section having the dimensionsas shown.

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EXAMPLE 7.8 (SOLN)

Shear-flow resultantsVertical downward shear V applied to sectioncauses shear to flow through the flanges and webas shown. This causes force resultants Ffand V inthe flanges and web.

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EXAMPLE 7.8 (SOLN)

Shear-flow resultantsX-sectional area than divided into 3 componentrectangles: a web and 2 flanges. Assume eachcomponent to be thin, then moment of inertia aboutthe neutral axis is

I= (1/12)th3

+ 2[bt(0.5h)2

] = (0.5th2

)[(h/6) + b]Thus, qat the arbitraryposition xis

q = =VQ

I

V(b x)

h[(h/6) + b]

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EXAMPLE 7.8 (SOLN)

Shear-flow resultantsHence,

The same result can be determined by first

finding (qmax)f, then determining triangular area0.5b(qmax)f= Ff

Ff= 0 q dx= =Vb2

2h[(h/6) + b]

b

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EXAMPLE 7.8 (SOLN)

Shear centerSumming moments about pt A, we require

Ve= Ffh

b2

[(h/3) + 2b]e =

As stated previously, edependsonly on the geometry of the x-

section.

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CHAPTER REVIEW

Transverse shear stress in beams is determinedindirectly by using the flexure formula and therelationship between moment and shear

(V = dM/dx). This result in the shear formula = VQ/It.

In particular, the value for Qis the moment of thearea Aabout the neutral axis. This area is theportion of the x-sectional area that is held on to

the beam above the thickness twhere is to bedetermined

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CHAPTER REVIEW

If the beam has a rectangularx-section, then theshear-stress distribution will be parabolic,obtaining a maximum value at the neutral axis

Fasteners, glues, or welds are used to connectthe composite parts of a built-up section. The

strength of these fasteners is determined fromthe shear flow, or force per unit length, that mustbe carried by the beam; q=VQ/I

If the beam has a thin-walled x-section then theshear flow throughout the x-section can bedetermined by using q= VQ/I

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CHAPTER REVIEW

The shear flow varies linearly along horizontalsegments and parabolically along inclined orvertical segments

Provided the shear stress distribution in eachelement of a thin-walled section is known, then,

using a balance of moments, the location of theshear center for the x-section can be determined.

When a load is applied to the member through

this pt, the member will bend, and not twist

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