SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY

DEPARTMENT OF CIVIL ENGINEERING

Subject:-Mechanics Of Solids

Topic:-Shear stresses on beam

Presented by:-Name Arvindsai

Dhaval Chavda

Fahim Patel

Navazhushen Patel

Enrollment no.130454106002

130454106001

140453106005

140453106008

Topics To Be Covered1. Shear Force2. Shear Stresses In Beams3. Horizontal Shear Stress4. Derivation Of Formula 5. Shear Stress Distribution Diagram6. Numericals

Shear forceAny force which tries to shear-off the

member, is termed as shear force.

Shear force is an unbalanced force, parallel to the cross-section, mostly vertical, but not always, either the right or left of the section.

Shear StressesTo resist the shear force, the element

will develop the resisting stresses, Which is known as Shear Stresses().

= = Shear force

Cross sectional area

SA

Example:-For the given figure if we want to

calculate the ..Then it will be Let shear force be S

=S/(bxd)d

b

S

Shear Stresses In Beams Shear stresses are usually maximum

at the neutral axis of a beam (always if the thickness is constant or if thickness at neutral axis is minimum for the cross section, such as for I-beam or T-beam ), but zero at the top and bottom of the cross section as normal stresses are max/min.

NANA

NA

When a beam is subjected to a loading, both bending moments, M, and shear forces, V, act on the cross section. Let us consider a beam of rectangular cross section. We can reasonably assume that the shear stresses τ act parallel to the shear force V.

v

n

V

z

m

O

b

h

Shear stresses on one side of an element are accompanied by shear stresses of equal magnitude acting on perpendicular faces of an element. Thus, there will be horizontal shear stresses between horizontal layers of the beam, as well as, Vertical shear stresses on the vertical cross section.

m

n

Horizontal Shear StressHorizontal shear stress occurs due

to the variation in bending moment along the length of beam.

Let us assume two sections PP' and QQ', which are 'dx' distance apart, carrying bending moment and shear forces 'M and S' and 'M+ ∆M and S+ ∆S‘ respectively as shown in Fig.

Let us consider an elemental cylinder P"Q" of area 'dA' between section PP' and QQ' . This cylinder is at distance 'y' from neutral axis.

F)(FQ'Fbe, shallcylinder in force

horizontal unbalancedHence,

dA δσσQ'F

dAx σp'F be, shallQ' and'P'at

stresses these to due forces The

xyIdMMδσσ

be, will'Q'at stress bendingSimilarly,

xyI

Mσ

be will P"at stress bending Hence,

This unbalanced horizontal force is resisted by the cylinder along its length in form of shear force. This shear force which acts along the surface of cylinder, parallel to the main axis of beam induces horizontal shear stress in beam.

Aax yI

dM

dA)(σd(σpFQFHF

DERIVATION OF FORMULA: SHEAR STRESS DISTRIBUTION ACROSS BEAM

SECTION Let us consider section PP' and QQ' as

previous. Let us determine magnitude of horizontal

shear stress at level 'AB' which is at distance YI form neutral axis.

The section above AA' can be assumed to be made up of numbers of elemental cylinder of area 'dA'. Then total unbalance horizontal force at level of' AS' shall be the summation of unbalanced horizontal forces of each cylinder.

a_yx

I

dMFM

dA.y1yy

1yy.X

I

dMdA.xy

1yy

1yy I

dMHF

Here, y = distance of centroid of area above AB from neutral axis, And a= area of section above AB.

This horizontal shear shall be resisted by shear area ABA'B‘ parallel to the Neutral plane. The horizontal resisting area here distance of centroid of area above AB from neutral axis and a=area of section above AB.

Ah = AB x AA’=b x dx where ‘b’is width of section at

AB.

We know that shear force is defined as S=dM/dx

Therefore, horizontal shear stress acting at any level across the cross sections.TH= Say / Ib

Ibyax

dxdM

dx.b

ayI

dM

xAF

forceshear horizontal resistingshear stressshear Horizontal

_

H

_

H

MH

SHEAR STRESS DISTRIBUTION DIAGRAM

1.Rectangular section

2.Circular section

maxNA

maxNA

3.Triangular section

4.Hollow circular section

h/2 max

avg

NA

maxNA

5.Hollow Rectangular section

6. “I” section

maxNA

maxNA

7. “C” section

8. “+” section

maxNA

maxNA

9. “H” section

10. “T” section

maxNA

maxNA

Numericals

Rectangular section sum Example-1: Two wooden pieces of a

section 100mm X100mm glued to gather to for m a beam cross section 100mm wide and 200mm deep. If the allowable shear stress at glued joint is 0.3 N/mm2 what is the shear force the section can carry ?

KN4FeShear forcKN4N4000F

10061067.6650810000F3.0

I.byFAτ

Νοω.

mm502

100Υ

2mm000,10100100A

joint. glued above beam ofpart consider

2N/mm3.0τ

4mm61067.66123200100I

:Solution

100mm

100mm

100mm

wooden piece

Circular section sumExample-2: A circular a beam of 100mm Diameter is subjected to a Shear force of 12kN, calculate The value of maximum shear Stress and draw the variation of shear stress along the Depth of the beam.

22.03N/mm

1.531.33

1.33Now.

21.53N/mm 7853.98

31012AF

27853.98mm21004πA

N3101212kNF

τmax

τavgτmax

τavg

:Solution

22.03N/mmτmax D

=100mmNA

I section sumExample-3: A rolled steel joist of I section overall 300 mm deep X 100mm wide has flange and web of 10 mm thickness. If permissible shear stress is limited to 100N/mm2, find the value of uniformly distributed load the section can carry over a simply supported span of 6m.

Sketch the shear stress distribution across the section giving value at the point of maximum shear force.

100mm

300mm

10mm

150mm

__

Y

10NA

25.63N/mminm 261.63N/mmavg

2100N/mmmax

mm10 b(givan)3N/mm100 τ

3mm257500

)7010140()14510110(yAN.A.at maximum be will stressshear

4mm61055.64Ixx

I 3xxI 2xxI 1xxIxx

4mm61029.180280012

328010I 2xx

4mm61013.232145110012

310110

2ahIgI 1xx

Solution:

kN/m56.83w

kN/mm64.836

36.501l

Wwu.d.l.

)load totalkN(36.501W2

W68.250

2WNow, F

kN68.250FN6.250679F

1061055.66257500F100

I.byFaτ

Triangular section sum Example-4: A beam of triangular section

having base width 150mm and height 200mm is subjected to a shear force of 20kN the value of maximum shear stress and draw shear stress distribution diagram.

2N/mm2

33.15.1 τave5.1τmax

2N/mm33.115000

31020AF

τave

2 mm1500020015021

bh21A

20kNF forceshear

Solution:

max=2N/mm2

avg=1.33N/mm2

200mm

150mm

h/22/3.hNA

Cross section sumExample-5: FIG Shows a beam cross section subjected to shearing force of 200kN. Determine the shearing stress at neutral axis and at a-a level. Sketch the shear stress distribution across the section.

50mm

100mm100mm 100mm

100mm

100mm

50mm50mm X

215.48N/mm 50610129.161005000310200

I.byFAτ

50mmb200kNF

100mm5050y

25000mm10050A

:line b_bat stressshear

4mm610129.6

]12

3100100[2]12

330050[xxI

:Solution

2N/mm03.5 250610129.16

812500310200

I.byFA τ

3mm812500

25)50(250100)100(50y A:axisx_x at stressShear

2mm/N09.32505015.48X

5.03N/mm2

15.48N/mm2

3.09N/mm2

100mm100mm 100mm

100mm

100mm

50mm50mm X

50mm

NA

Inverted T section sumExample-6: Shows the cross section of a beam which is subjected to a vertical shearing force of 12kN.find the ratio the maximum shear stress to the mean shear stress.

60mm

20mm

60mm

20mm

46

23

23

xx2xx1

21

2211

22

2

12

1

mean

max

3

mm101.36

30)1200(501212

602010)(30120012

20 60

II I inertia ofMoment

30mm 12001200

50 1200 10 1200AA

yAyA y

50mm202

60y 1200mm6020 A

10mmy 1200mm2060 A AxisNeutral of Position

ττ:find To

N101212kNs :Given:Solution

2.21ττ

511.029

ττ

stressshear mean to stressshear maximum of Ratio

5MPa2400

1012area c/s

forceShear τ

stressshear Mean

11.029MPa101.3620

2510001050 τ

axis neutral above areaConsider bI

SAy τ

axis, neutralat produced stressshear Maximum

mean

max

mean

max

3

mean

6

3

max

max=11.29MPA

avg=5 MPAmin =2.21 MPA

20mm

60mm

20mm

NA

L section sum Example-7: An L section 10mm X 2mm show in the fig.

is subjected to a shear force F. Find the value Of shear force F if max. shear stress developed is 5N/mm2.

4208.09mm

24)(6.771612

382

2ahgIxx2I

4106.12mm26.77)(92012

3210

2ahgIxxI

6.77mm1620

416920

2a1a2y2a1y1a

y

4mm2y

216mm282a

9mm1y

220mm2 101a

:Solution

2mm

2mm

10mm

10mm

6.77mm __Y

3.23mm

68.14NF 2314.21

46.11F5

I.byFA τ

46.11mm 2

1.232)(1.232.232)(10yA

N.A.at occur will stressshear Maximum

314.21mm 208.09106.12

III

3

4

xx2xx1xx

Tee section sum Example-8: A beam is having and subjected to

load as shown in fig. Draw shear stress distribution diagram across the section at point of maximum shear force, indication value at all important points.

100kN

A B

3m 3m

100200 200

300

100

25

275mm __Y

K.N 502100F

4mm 3750000275)-(350(500x100)xy A

: web & flange of junction theat stressShear

4mm 91.02x10

6693.57x106322.91x10 12

2150)-100)x(275x (500 3100x(300)

12 275)-(350x (500x100) 3500x(100)I

mm 275 80000

612x10

100)x (300100)x (500 100)x15x (300100)x350x (500Y

XX

2

29max

3

2avg

29min

N/mm85.1 500x10x02.1

3781250x3)10x(50 I.b

yFA τ

mm3781250 )5.12x25x100()75x100x500 (yA

axis, neutralat stressShear

mm N/84.1 100

500x367.0 τ

to, increase suddenly will stressShear

mmN/367.0 500x10x02.1

3750000x3)10x(50 I.b

yFA τ

mm500b

max=1.85N/mm2

avg=1.84N/mm2

min=0.367N/mm2100200 200

300

100

25

275mm __Y

NA

I section sum Example-9: Find the shear stress at the

junction of the flange and web of an I section shown in fig. If it is subjected to a shear force of 20 kN.

100mm

200mm

20mm

100mm

20

2

26

3

3_

463

xx

mm/N57.4915.0x2

100 =

to, increased suddenly will stressShear

mm915.0100x10x36.39

180000x10x20b.I

_yAF

mm18000090x)20x100(yA

:web and flang of junetionat streesShear

mm10x36.3912

200x100I

N.k20F

:Solution

100mm

200mm

20mm

100mm

20NA

2N/mm 0.915τmin

2N/mm915.0avg

2N/mm75.4max

Rectangular section sum Example-10: A 50mm x l00mm in depth

rectangular section of a beam is s/s at the ends with 2m span the beam is loaded with 20 kN point load at o.5m from R.H.S. Calculate the maximum shearing stress in the beam.

20kN

RB

B

0.5m

2.0mR

A

A

2N/mm5.4

00.35.1 τave5.1τmax

2N/mm00.3100x5031015

AF

τave

kN51520B

B

B

RkN15R

5.1x202xRat Amoment Taking

Solution:

max=4.5N/mm2

50mm

100mm NA