chapter 7_tranverse shear

7/27/2019 Chapter 7_Tranverse Shear

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Russell C. Hibbeler

Chapter 7: Transverse Shear


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Shear Stress In Beams

Objectives:

To develop a method for finding the shear stress in abeam.

having a prismatic cross section

made from homogeneous material

behaves in a elastic manner


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The shear V is the result of atransverse shear-stress distributionthat acts over the beams crosssection.

Due to the complementary propertyof shear, associated longitudinalshear stress will also act alonglongitudinal planes of the beam


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000

00

x z xz z

x y xy y xy xz x x x

y M dA F

dA z M V dA F

dA z y M dA F

Distribution of normal and shearingstresses satisfies

Transverse loading applied to a beamresults in normal and shearing stresses intransverse sections.

When shearing stresses are exerted on thevertical faces of an element, equal stressesmust be exerted on the horizontal faces

Longitudinal shearing stresses must existin any member subjected to transverse

loading.


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Physically illustrate why shear stress develops on the longitudinal planes

If the top and bottom surfaces of eachboard are smooth, then the applicationof the load P will cause the boards toslide relative to one another . The beamwill deflect as shown.

The longitudinal shear stressesbetween the boards will prevent therelative sliding of the boards andconsequently the beam will act as asingle unit.


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As a result of the shear stress, shear strain will be

developed and these will tend to distort the cross section!!!

When a shear V is applied, it tends todeform the lines into the patternshown. This non-uniform shear-straindistributed over the cross section willcause the cross section to warp (notto remain plane)


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Shear FormulaThe shear formula was derive by considering horizontal force equilibrium of the

longitudinal shear stress and bending stress distributions acting on a portion of adifferential segment of the beam.


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Consider the shaded top segment


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It

VQ

= the shear stress in the member at the point located a distance yfrom the NA. This stress assume to be constant .

V = the internal resultant shear force.I = the moment of inertia of the entire cross-sectional area (about theNA)

t = the width of the members cross-sectional area, measures at the

point where

is to be determinedQ = yA,where Ais the top (or bottom) portion of the members cross-sectional area, define from the section where t is measured, and y isthe distance to the centroid of A,measured from the NA


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Shear Stress in Beams

Typical shear failure


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Maximum Shear Stress For A Beam With Rectangular Cross Section

V

bc yc

bbh

ycbV

It VQ

ycb

yc ycb y AQ

pq

area shaded

322

3

22

22

4

3

12

2

22

1''

If the cross sectional area is A = b(2c ),

2

2

2

22

12

3

22

3

c y

AV

V cbc yc

pq

For layers, y = c:

012

32

2

cc

AV

rs

For layer at neutral axis, y = 0:

A

V

c A

V NA 2

301

2

3

2m ax

AV

23

max


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Maximum Shear Stress For A Beam With Circular Cross Section

r

3

4 r

m ax

NA

464

44 r d I

3

2

3

4

2

''

3

2

r

r r

y AQ

A

V r V

r r

r V It

VQ

3

4

3

4

24

3

2

2

4

3

m ax

AV

34

max


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Maximum Shear Stress For A Beam With Hollow Circular Cross Section

444 io

r r I

33

22

3

2

34

234

2

''

io

iioo

r r

r r r r

y AQ

22

22

44

33

max

3

4

24

3

2

io

iioo

ioio

io

r r

r r r r

A

V

r r r r

r r V

It VQ

ma x

NA

r r

oi

ma x

22

22

max 34

io

iioo

r r r r r r

AV


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The beam shown below is made of wood and is subjected to a resultantinternal vertical shear force of V = 3 kN. (a) Determine the shear stress inthe beam at point P, and (b) compute the maximum shear stress in thebeam.

Example 1


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3 mm ..' 4107518100505021512 A yQ

4633 mm1028.16125100121

121

bh I

(a) The moment of inertia of the cross sectional areacomputed about the neutral axis is

Applying the shear formula, we have

(Ans) MPa346.01001028.161075.183

6

4

It VQ

p


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(b) Maximum shear stress occurs at the neutral axis

(Ans) MPa360.01001028.16

1053.1936

4

max It VQ

34 mm1053.195.621002

2.65''

A yQ

Applying the shear formula yields


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Example 2The beam is fabricated from three steel plates, and it is subjected to ashear force of V = 150 kN. Determine the shear stress at point A and Cwhere the plates are joined. Show y = 0.080196 m from the bottom and I NA = 4.8646 x 10 -6 m 4.


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yC = 32.3 mm

ANA

CB

yB = 42.696 mm

100 mm

10 mm

15 mm10 mm 100 mm 10 mm

65 mm

y = 80.196 mm 75 mm


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The beam shown below is made from two boards. Determine themaximum shear stress in the glue necessary to hold the boards together along the seam where they are joined. The supports at B and C exert onlyvertical reactions on the beam.

Example 3


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m12.0

15.003.00.030.15075.015.003.00.1650.030.15

y

46

23

23

1027

12.0075.015.003.012

15.003.0

12.0165.003.015.012

03.015.0

m

I NAkN5.19maxV

33102025.0

015.012.018.003.015.0

''

m

y AQ D


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MPa88.403.01027 102025.0105.19 633

It VQ


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Example 4Determine:a. The shear stress at point B on the web of the cantilevered strut at

section a a.b. The maximum shear stress acting at section a - a

2 kN 4 kN


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Example 6

The T-beam is subjected to the loading shown. Determine the maximumshear stress in the beam at the critical section.


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2008 Pearson Education South Asia Pte Ltd


Mechanics of Material 7 th Edition

Shear Flow in Built-Up Members

For fasteners it is necessary to know the shear force by the fastener along the members length .This loading is referred as the shear f low q , measured as a forceper unit length.

I VQ

qq = shear flowV = internal resultant shear I = moment of inertia of the entire cross-sectional area


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SHEAR FLOW IN BUILT-UP MEMBERS

Shear flow : Is a measure of the force per unitlength along a longitudinal axis of beam. Found from shear formula and is used todetermine the shear force developed in fasteners

(nails, bolts) and glue that holds the varioussegments of the beam.

The magnitude of shear flow can be obtainedusing development similar to that finding the shear

stress in the beam.


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In practice, members arebuilt from several compositeparts in order to achieve a

greater resistance to loads.

Fasteners may beneeded to keep the

component parts fromsliding relative to oneanother. In order toknow the shear forcethat must be resisted

by the fastener alongthe members length


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Three horizontal forces act on this part.

1. Two forces, F and F + dF , are developedby normal stress caused by the momentsM and M + dM , respectively

2. For equilibrium, d F acts at the juncture andis to be supported by the fastener.

I d M

d F = ydA A

q = dF/dx and V = dM/dx

I

VQ q =

Shear Flow


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q = the shear flow, measured as force per unit length along the beam.

V = the internal resultant shear force.

I = the moment of inertia of the entire cross-sectional area (about the NA)

Q = ydA = yA , where Ais the cross-sectional area of the segment thatis connected to the beam at the juncture where the shear flow is to becalculated, and y is the distance from the NA to the centroid of A.

I

VQ q =


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Value of q

Single Fastener,

Fig. (a) and (b)

Two Fasteners ,

Fig. (c).

Each fastener supports q/2

Three Fasteners,

Fig. (d).

Each fastener supports q/3


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Example 7

A box is to be constructed fromfour boards together as shown. If each nail can support a shear force of 30 N, determine themaximum spacing S of the nails atB and at C that the beam willsupport the vertical force of 80 N.


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Example 8The beam is constructed from threeboards. Determine the maximumshear V that it can sustain if the

allowable shear stress for the wood is allow = 4 MPa. What is the requiredspacing S of the nails if each canresist a shear force of 800 N.


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Example 9The strut is constructed from threepieces of plastic that are gluedtogether as shown. If the allowableshear stress for the plastic is allow = 8 MPa and each glue joint can

withstand 250 N/cm, determinethe largest allowable distributedloading w that can be applied tothe strut.

If the distributed load w = 2000N/m, determine the shear stressthat must be resisted by each glue

joint.


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2008 Pearson Education South Asia Pte Ltd


Mechanics of Material 7 th Edition

Example 10The beam is constructed from four boards glued together. If it is subjectedto a shear of V = 850 kN, determine the shear flow at B and C that mustbe resisted by the glue.


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46 m1052.87 I

m1968.0~

A

A y y

The neutral axis (centroid) will be located from the bottom of the beam,

The moment of inertia computed about theneutral axis is thus

33 m10271.001.0250.01968.0305.0'' B B B A yQ

Since the glue at B and holds the top board to the beam


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MN/m0996.01052.87

1001026.0850'

MN/m63.2

1052.87

10271.0850'

6

3

6

3

I VQ

q

I

VQq

C C

B B

(Ans) MN/m0498.0 and MN/m31.1 C B qq

Likewise, the glue at C and C holds the inner board to the beam

Therefore the shear flow for BB and CC,

33

m1001026.001.0125.01968.0205.0'' C C C A yQ

Since two seams are used to secure each board, the glue per meter length of beam at each seam must be strong enough to resist one-half of each calculatedvalue of q.

chapter 7_tranverse shear

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