Strength of Materials- Shear Stress in Beams- Hani Aziz Ameen

Strength of Materials Handout No.9

Shear Stress in Beams Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]

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Strength of materials- Handout No.9 - Shear Stress in Beams- Dr. Hani Aziz Ameen

9-1 Introduction When a shearing force is applied to a beam , the shear stress on the

cross section produces sliding of transverse elements of the beam . The mean shear stress on a transverse section is the shearing force divided by the cross sectional area , but this stress is not uniform a cross the section , being zero at the top and bottom of the section . It may also vary across the width of the section but in the following analysis , the shear stress is assumed to be uniform on planes parallel with the neutral axis . 9-2 Derivation of Shear Stress Formula

Fig(9-1) From Fig(9-1a) ,

0Fx dF = H2 H1

c

y1

c

y2

11

dA dA dF

substitute I/My

c

y

c

y

12

1 1

dA I

yMdAI

y MdF


c

y

c

y

12c

y

12

1 11

dA yI

MMdA yI

MdAy I

MdF

We have dF = bdx and the change M2 M1 represents dM at dx Hence ,

c

y1

ydAI

dMbdx

c

y1

ydAbdx IdM ............... (9-1)

Now from the Fig(9-2a)

Fig(9-2) MB= 0

M+Vdx ( M+ dM) = 0 V dx dM = 0 V= dM / dx where V is the vertical shear force Hence , eq. (9-1) , will be

c

y1

ydAb I

V

y Ab I

V . (9-2)


A is the area of the shear section y is the distance from the centre of shear section area to N.A.

9-3 Variation of Shear Stress in Beam Cross Section

1- For rectangular cross section From Fig(9-3a)

-a- -b- Fig(9-3)

y Ab I

V

h)h2d(

21*)h

2d(b

bV

12bd3

h2h

2d

21*)h

2d(b

bV

12bd 3

2

2

33 h4

dbd

V6h2d

21*)h

2d(

bdV12

2

2

3 h4

dbd

V6


This is a parabolic distribution as shown in Fig(9-3b) . The max. ordinate

being bd2V3 when h = 0 , thus the max. shear stress is (3/2) times the

mean shear stress .

2- For circular section From Fig.(9-4)

Fig(9-4)

y Ab I

V = r

h

ydAIBV

dy yr2 yIBV r

h

22

2/322

2244

)hr(32

hr2r

V

)hr(rV4 22

4

This is again a parabolic distribution as shown the max. ordinate being

2rV

34 when h = 0

Thus the max. shear stress is (4/3) times the mean shear stress .


3 For symmetrical I section A Shear Stress in Flanges

y AIbV

yA = b ( ( D/2) y) * ( (D/4) (y/2) + y) yA = b ( D/2) y) * 0.5* ((D/2)+y) = (b/2)*( (D2/4) y2)

22

22

y4

DI2

V)y4

D(2b*

IbV

B Shear Stress in Web

y AIbV

y A = b * ( (D d)/2)*((D+d)/4)+((d/2) y)b * 0.5*((d/2)+y)

y A =(b/8) (D2 d2) + (b /2)( (d2/4) y2)

)]y4d(b)dD(b[b I 8

V 2222

max at y = 0

)dD(bb I 8

V 22max


4- For Other Typical Section

L- section

Hollow circle

T-section

cross

Built up section


9-4 Spacing of Connecting Elements in Built up Beams A beam may be formed from a number of pieces of a certain

material by riveting or bolting these by pieces together for the purpose of increasing the moment of inertia of the beam . Such a beam is called a built up beam . Large girders are usually built up by welding or riveting the flange plates to the web . The weld or rivets must therefore transfer the horizontal shear force between the web and flange .

-a- -b- Fig(9-5) In order that for the built up beam to function as a beam of one piece , we must determine the necessary spacing of the connecting elements . To do this , let us refer to the cantilever beam of Fig(9-5 a) which is composed of four wooden planks bolted together and loaded with a concentrated load P at its free end From Fig(9-5 b) , neglecting friction between the contact surfaces of the planks , the difference are normal farces ( F2 F1 ) [ which can be formulated from A*b ]

From Fig.(9-5b) , in order to satisfy the equilibrium condition : F1 + H F2 = 0 . 9-3) H = F2 F1

can be get H = * A H = * e* b 9-4 ) The shearing stress at the surface of contact = (V/ IB) * yA Eq. (9-4) will be H = (V / Ib )* yA . e. b H = (( V. e ) / I ) yA where :

A Fig.(9-4b) y Distance from the centroid of the shaded area to the neutral axis .

Moment of inertia of entire section a round the neutral axis .



9-5 Examples The following examples exlain the different ideas on the shearing stress problems. Example(9-1) Fig(9-6) shows a simply support beam with rectangular cross section of b = 120 mm , h = 180 mm , and its length of 6 m , carrying a uniform distributed load of 4kN/m .Find :

(a) The shear stress in the fiber at 30 mm from the extreme to the cross section away 1 m from the left side .

(b) The max. shear stress in the beam .

Fig(9-6) Solution

a) I = bh3/12 = (120*(180)3) /12 = 58.32*10 6 m4

yAIbV

y = 75 mm = 0.075 m A = 30*120 mm2 = 0.12*0.03 m2 I = 58.32 * 10 6 m4 b=0.12 m To find V ( vertical shear force)


Fy = 0 , R1+R2 = 4*6 = 24

M =0 , R2*6 = 24*3 , R1 = 12 kN R2 = 12 kN 0 x<6 Fx = R1 4x At x = 0 Fx = R1 = 12 kN At x = 6 Fx = 12 4*6 = 12 24 = 12 kN Fx at x = 1 m = R1 4 *1 = 12 4 = 8 kN

= ( 8*1000)*(0.12*0.03)*(0.075) / ( 58.32*10 6*0.12) = 309 kPa

b) yAIbV

V = 12 kN I = 58.32*10 6 m4 b = 0.12 m A =90*120 mm2 = 0.09 * 0.12 m2 y =45 mm = 0.045 m

= ((12*1000)*(0.09*0.12)* 0.045 )/(58.32*10 6*0.12)= 833.33kPa Example(9-2) Fig(9-7) shows I- wide flange cross section of the beam which has V = 70 kN , find :

a) Max. shearing stress b) Shear stress at the inter section point of web & flange c) Plot the shear stress distribution for the section d) Determine the percentage of shear force in the web .


Fig(9-7) Solution

yAIbV

Iz = I rect (1) 2* I rect(2) = (bh3/12) rect(1) 2 (bh3/12)rect(2) Iz = ( 160* (240)3)/12 2*(70*(200)3/12)= 91*106 mm4

A y = (160*20)*110+(100*20)*50 b = 20 mm = [(70*103)/(91*10 6*0.02)] [ (0.16*0.02*0.11)+(0.02*0.1*0.05)] = 17.4 MPa At the intersection point , the width b will be b = 160 mm at the flange b = 20 mm at the web

y AIbV = [(70*103)/ ((91*10 6)*0.2)]*(0.16*0.02)*(0.11)=13.5MPa


The average of parabolic = (2/3)* (17.4 13.5) = 2.6 MPa av. = the average of the shear stress in the web = 13.5 + 2.6 = 16.1 MPa

Shear force in the web = A av. = (20*200)*10 6*16.1*106=64.4 kN The percentage shear force on the web = (64.4/70) *100 = 92% Example(9-3) Prove that the max. shear stress of the pipe with cross section area A is

= 2V/A Solution

A=2* * r * t I N.A. = 0.5 A 2

y AIbV = r2

2A

)t2(AV2

21

AV2

2/AV

rtV


Example(9-4) Fig(9-8) shows a simply supported beam of 4 m length with the cross section . Find the uniform distribution load applied to beam if =1.2 MPa

Fig(9-8) Solution INA= [(150*2003)/12] [(100*1503)/12] = 71.88 * 106 mm4

yA =(150*100*50) (100*75*37.5) yA = 468.8*10 6 m3

y AIbV ,

y AIbV

Now to find the shear force , we draw the S.F.D. 0 x<4 Fx = R1 wx At x =0 R1 = 2w At x = 4 Fx = 2w 4w = 2 w V = 2w 2w = ( 1.2*106*71.88*106*0.05) /(468.8*10 6) w = 4.6 kN/m


Example(9-5) Fig(9-9) shows a beam with triangle cross section. If V is the shear force, prove that the max. shear stress = (3V/ bh) and locate its position

Fig(9-9) Solution

y AIbV

= (V/Ix) * ( 0.5 xy)[(( 2/3) y) + (y/3)] = ( V / 3 I) *( h y y2)

d / dy = ( V/ 3 I ) * ( h 2 y) = 0 y = h/2

bhV3

4h

2h

3V 22

36bhmax 3

Example(9-6) Prove that for the cross section beam shown in Fig(9-10) that the max. shearing stress occurs at ( h/8 ) . Solution

y AIbV

= (V/Ix) * ( 0.5 xy [(h/2) (2y/3)]

= ( V / 2 I) *( h y/2 2y2/3) d / dy = ( V/ 2 I ) * ( h/2 4 y/3) = 0 y = 3h/8 from top or h/8 from N.A. Fig(9-10)


Example(9-7) Fig(9-11) shows a wide flange simple beam supports a concentrated load of 53.37kN and a uniformly distributed load including its own weight of 29.18 kN/m .Find the shearing stresses acting at the points indicated on the beam . also , find the normal stresses acting at the points D & E .

Fig(9-11) Solution

Fy = 0 R1+R2 = 53.37 + 29.18*7.31 = 266.88 kN MH= 0 R1 * 7.31 53.37*3.657 29.18*7.31*3.657 =0 R1 = 133.44 kN , R2 = 133.44 kN 0 x1<3.667 Fx1 = R1 29.18 x1 Mx1 = R1 x1 29.18 x1 x1/2 At x1 = 0 Fx1 = R1 = 133.44 kN Mx1 = 0 At x1 =3.657 Fx1 = 133.44 29.18 * 3.657 = 26.68 kN Mx1 = 133.44 * 3.657 29.18 *(3.657)2 /2 = 292.8 kN.m 0 x2<3.657 Fx2 = R1 29.18 *3.657 53.37 29.18*x2 Mx1 = R1 (x2 +3.657) (3.657*29.18)(1.82+x2) 53.37 x2 29.18 x2 x2/2 At x2 =0 Fx2 = R1 3.657*29.18 53.37=26.688 kN Mx2 = 133.44*3.657-(29.18*3.657) * 1.82= 292.8 kN.m At x2 = 3.657 Fx2 = 133.44 3.657*29.18 53.37 29.18*3.657 =133.44 kN


Mx2 =133.44*7.315-3.657*29.18*5.48-53.37*3.657-29.18-(3.657)2/2 =0

S.F & B.M at x2 = 1.82 m Fx2 = 133.44 3.657*29.18 53.37 29.18*3.657 = 80 kN Mx2 = 133.44*(1.82+3.657) (3.657*29.18)(1.82+1.82) 53.37*1.82 (29.18/2)*(1.82)2 = 195.21 kN

Iz = Iz)I 2 Iz)II Iz = 729.4 x 106 mm4 Level 1 1 : point A

y AIbV

T =0 level 2-2 , point B A =228.6* 20.193 = 4616.1 mm2 y = 256.54 mm b = 228.6 mm


= (80000*4616.1x10-6*256.54x10-3) / (729.43x10-6*228.6x10-3) =572.285 N/m2 Level 2-2 , point C A = 228.6* 20.193 = 4616.1 mm2 y = 256.54 mm b = 12.67 mm

= (80000*4616.1x10-6*256.54x10-3) / (729.4*10-6*12.67x10-3) =10.27 Mpa Level 3-3 , point D

y A = [(20.193*10-3*226.6*10-3)*(266.7*10-3 (20.193/2))] +[(266.7*10-3 20.193) *12.67*10-3]((266.7*10-3 20.193)/2) = 1.57 mm3 b = 12.67 mm

= (80000*1.57*10-9) / (729.4*10-6*12.67*10-3) = 13.652 MPa Level 4-4 , point E

y A = [(20.193*226.6*10-3)*(266.7*10-3 (20.193*10-3/2))] +[(266.7*10-3 20.193*10-3) * 12.67*10-3]*((266.7*10-3

20.193*10-3)/2) + 12.67*10-3 ((266.7*10-3 152.2*10-3)/2) = 0.766*106 mm3 b = 12.67 mm

= (80000*0.766*10-3) / (729.4*10-4*12.67*10-3) = 12.82 MPa Normal stress at D & E

= M y / I at point D , y = 0 , at D = 0 at point E

= M y / I = [(195.2*10-3) * (266*10-3 152.2*10-3) ] / 729.4*10-6 = 30.68 MPa Example(9-8) Fig(9-12) shows a simply supported beam which carries a uniformly distributed load of w N/m and 2w N/m .If the beam is made of wood and has the section shown in Fig. Find the max. distribution loads in N/m that can be supported by the beam without exceeding a shearing stress of 0.689 MPa and a bending stress of 6.895 Mpa .


Fig. (9-12) Solution

Fy = 0 R1 + R2 = 1.2192* 2 w + w* 0.6096 = 3.048 w MB = 0 1.8288 R1 ( 2w*1.2192)*(1.2192)

0.6096 w * 0.3048 = 0 R1 = 25.2 w R2 = 19.27 w 0 x1<1.2192 Fx1 = R1 2 w x1 Mx1 = R1 x1 2 w x1 x1/2 At x1 = 0 Fx1 = R1 = 25.2 w , Mx1 = 0 At x1 = 1.2192 Fx1 = 25.2 w 2 w *(1.2192)2/2 = 10.37 w Mx1 = 25.2 w *1.219 2 w (1.21922/2) = 9.037 w 0 x2<0.6096 Fx2 = R1 2 w *1.2192 w x2 Mx2 = R1 (1.2192+x2) ( 2w*1.2192) (0.6096+x2) wx2 x2/2 At x2 = 0 Fx2 = 25.2 w 2*1.2192 w = 10.37 w N Mx2 = 25.2 w *1.2192 2*1.2192 w*2 =9.037 w At x2 = 0.6096 Fx2 = 25.2 w 2*1.2192 w 0.6096 w = 19.27 w Mx2 = 25.2 * 1.8288 w 2*1.2192 w * 1.2192 w (0.60962/2) = 0


y AIbV

max

4hb

2h

b

w2.2510*689.012

3bh

6

w = 13.7 kN/m To find Mmax dMx1/ dx1 = R1 ((2w/2) 2x1) =0 x1 = R1/2w = (25.2w) (1/2w) Mmax = 25.2 w* 0.863 (2w/2) *(0.63)2

= (103/12) w I = bh3/12 = (203.2*2543)/12 ymax = 127 mm

I/My

1212

3254*2.203

26

10*

127.0*)63.0(w863.0*w2.2510*895.6

w = 18.883 kN/m


Example(9-9) Fig.(9-13a) shows the simply supported beam is composed of three wooden planks screwed together so as to form the cross section in Fig(9-13b). If the beam carries a uniformly distributed load of 5.837kN/m over its entire length, Find the spacing of the screws in the built up beam, Assume that each screw can resist a total force of 1.668 kN.

Fig(9-13) Solution In order to find the spacing e , we have to locate the position of the neutral axis on the cross section of the beam This is found as follows

AyAy

mm2542.76*4.1522.76*8.3042.76*254

1.38*2.76*4.152)2.764.152(*2.76*8.3041.419*2.76*254y1

and 2y = 457.2 254 = 203.2 mm IN.A. = [(152.4*2543)/3] [(152.4 76.2)*(254 76.2)3/3]+ (254*203.23/3 ) [(254 76.2)*(203.2 76.2)3/3] = 1278.6*106 mm4 Upper flange

eI

y AVH

Fy= 0 , R1+R2 = 5.837*6.096

MB = 0 , 6.096*R1 5.837*6.096*3.048 = 0 R1 = 17.79 kN R2 = 17.79 kN


0 x<6.096 Fx1 = R1 5.837x1 At x1 = 0 Fx1= R1= 17.79 kN At x1 = 6.096 Fx1 = R1 5.837*6.096 = 17.79 kN

V= 17.79 kN

A = area of upper flange = 152.4*76.2 =11612.8 mm2 y = distance from the centroid of the area of the upper flange the Neutral axis = 254 38.1 = 215.9 mm H = 1.668 kN

1.667*103 = e10*6.1278

10*9.215*10*8.11612*10*79.176

363

e =47.75mm Lower flange A = 76.2*254 = 762mm2 , y = 203.2 127 = 165.1 mm

1.66*103 = 6

663

10*6.127810*1.165*10*762*10*79.17 e

e = 37.338 mm Hence , the screws are spaced at intervals of 47.75mm , on the upper flange and at 37.33mm on the lower flange .


Example(9-10) Fig(9-14) shows a T section beam , symmetrical about a vertical axis is made with a top flange 100 mm wide and 14 mm thick to which a vertical web plate , 150 mm deep and 10 mm wide is welded . At a certain section, the total shearing force is 40 kN. Find the percentage shear carried by the vertical web and the shearing force per meter run in the welded connection .

Fig(9-14) Solution Taking moment about the Top section 100 *14*7+15*10*89 = (100*14+150*10)*h h = 49.4 mm

IN.A.= 12

14*100 3+100*14*42.42 +

12150*10 3

+10*150*39.62

IN.A.= 7.708*106 mm4 A distance x m from the bottom of the web

))2/x(1146.0(*01.0*01.0*10*708.7

10*406

3

=2593 x (0.2292 x) MPa Shear force carried by element = 2593*(0.2292 x) * 0.01 dx

Total shear force carried by web = 25.9315.0

0

2 dx)xx2292.0(

= 0.0377 MN Percentage of shear force carried by web = (37.7/40)*100 = 94.2 Shear stress at top of web = 2593*0.15*(0.2292 0.15) = 30.8MPa Shear force per meter in welded connection = 30.8*106*0.01 * 1 = 308 kN


9-6 Shear Center

9-6-1 Shear Flow in Beams Having Thin Walled Section

The shearing stresses in beams are computed by

y AIbV as discussed before in Article 9-2 (Eq.(9-2))

Consider a wide flange or an I beam fixed at one end and subjected to a vertical shearing force P at the other free end as shown in Fig(9-15).

Fig(9-15) Applying equilibrium equation on the element . N2 = N1 + dN dN= N2 N1 9-5) and N2 = stress * area = 2*(t * a ) N1 = 1 * ( t * a ) N2 N1 = t * a ( 2 1) we have bending load , thus = ( M y ) / I N2 N1 = t * a [ (((M + dM) * y )/I) (M y )/I ] dN = t * a ( dM * y / I ) (9-6)


and the longitudinal shearing stress force dN is dN = shearing stress * area dN = * ( t* dx) 9-7) from Eq.(9-6) & Eq.(9-7)

*( t * dx) = t * a * dM * ( y /I)

* t = t * a * ( dM/dx) * ( y /I) where

q = ( t* a) ( F. y . (9-8) The shear flows in the flange at any distance ( a ) from the edge of the flange , therefore the shear flow in the flanges is a max. when the distance ( a ) is equal to (0.5 * length of one flange) Hence , q = ( t* b) (F y /I) , where b= 0.5 * length of one flange ( t * b) = Af = 0.5 * area of one flange q = F Af y /I ....... (9-9) Notice that in calculating the max. shear flow for the web by the formula q = F Af y /I , the area A is either that above or below the neutral axis of the cross section and y is the distance from the neutral axis to the centroid of the area A ; the max. shear flow in the web occurs at the neutral axis of the section . An alternative way of finding the maximum shear flow in the web of finding the maximum shear flow in the web is to proceed as follows. The max. shear flow in the web is that of the upper or lower flange plus the shear flow from the neutral axis to the center of the upper or lower flange . Thus , qweb = 2 q + F A1( y 1/I )


A1 upper or lower flange y 1 is the distance from the neutral axis to the centroid of A1 qweb = 2 F Af ( y /I) + F A1( y 1/I) The horizontal shearing force dH acts only on the area ( t. a) of the flange, therefore the horizontal shearing force H acting on each half of the flange is computed from the relation H = average shear flow * ( ½ * length of one flange) H = ½ q*b H = ½ * F Af ( y /I) * b ................... (9-10) where b is the distance from the edge of the flange to the center line of the web

9-6-2 Shear Center The shear center is defined as the point on the cross section of the beam through which the vertical shearing force ( P)must pass in order to avoid twisting as the beam bends . One of the necessary requirement in the application of the flexure formula =M y/I is that the beam must not twist as it bends . Bending without twisting can be accomplished only when the applied vertical shearing force ( P ) passes through the shear center or center of twist


Consider a cantilever beam having a channel section as shown in Fig(9-16)

Fig(9-16) Shear center for the cannel section is located from the relation

Mo= 0 H * 2 y 9-11 ) From Eq.(9-10) , it can be given H * 2 * y = F * e ½ F Af ( y /I) * b ( 2 y ) = F * e y = ½ h , Af=b . t e = ( h2 b2 t ) / ( 4 * I ) where

is the vertical depth between the center lines of the flanges is the distance from the center line of the web to the shear center O

will not twist as it bends .

9-6-3 Shear Center of Unequal Flanges In the H section of unequal flanges of Fig(9-17) the neutral axis is the only axis of symmetry and the shear center O is not coincident with the centroid C of the cross section . The shear center O for such a section , neglecting the bending resistance offered by the web , is located as follows : Referring to the cross section of Fig.(9-17) and applying the relation

Mo= 0 , we obtain


Fig(9-17) P = F1+F2 F1* e1 = F2 * e2 F1 & F2 In bending , the two flanges will have the same raduis of curvature , hence ; ( 1/R) = ( M1/E I1) = ( M2 / E I2) ( M1/I1) = ( M2/I2) M1 = F1 * h M2 = F2 * h ( F1 / I1) = ( F2/I2) and ( I1 / I2) = ( e2 / e1) The total moment of inertia I = I1+I2 and h= e1+e2 ( I I2) / I2 = ( h e1) /e1 It can be obtained ( I / I2) = ( h / e1) or e1= h (I2/I) e2 = h(I1/I)


9-7 Examples The following examples explain the different ideas of the shear center . Example(9-11) Fig(9-18a) shows the cross-section of a cantilever beam carries a downward vertical load of 3336 N . Assuming the thickness t =2.54 mm . of the wall is uniform through the section, construct a shear flow diagram and locate the shear center for the section .

Fig(9-18a) Solution The moment of inertia of the cross- section around the neutral axis is I = 2*{ 2.54*76.2*1272+2.54*50.8*76.22 + (2.54*1273)/3 } =11.196*106mm4 Notice that the area of a flange may be considered as a strip and there fore its moment of inertia about the neutral axis is given by the product of the area and the square of the distance from the centroid of the area to the neutral axis. Shear flow diagram : Flanges A and B For the Flange A or B, A y = 2.54 * 76.2 * 127 = 24.58*103 mm3

Applying equation ( 9-9) , q =FAf y /I we have q = 3336*24.58*103*10 9 / 11.196*10 6 = 7319.93 N/m


The horizontal shearing force acting on the flange A or B is computed

H = q (average ) * b where b = 76.2 mm .

H = ½ * 7319.93*76.2*10 3 = 1592.58 N Shear Flow diagram : Flanges c & d Following the above procedure

A y = 2.54*50.8*76.2 = 9832.23 mm3 q =3336*9832.23*10 9 / 11.196*10 6 =2933.2 N/m and H =(1/2)*2933.2*50.8*10 3 = 425.45 N The shear flow diagrams for the flanges A , B ,C and D and the horizontal shearing Forces acting on the flanges are shown respectively. In Fig (9-18 b&d). It is to be noted that the shear flow or shearing stress is always zero at a free surface, such free surfaces are the edges A,B,C,& D , as can be seen from equation q = (t a) F y / I thus, when the distance along the flange is zero , the shear flow or shearing stress is also zero .


Fig(9-18 b,c,d)


Shear flow diagram : web

The shear flow at the level 1 or 2 of Fig(9-18 a) is computed by the equation q = FA y / I . Just before level 1 or 2 is reached , we have A y =area of the section above level 1 or below level 2 . Multiplied by its y = 2.54* 76.2*12.7 +2.54*50.8*(25.4+76.2) = 37.690*103 mm3 Hence, q (just before level 1 or 2 )= (3336*37.690*103)/11.196*10 6 = 11.207 kN/m Alternatively , the shear flow just before level 1 or 2 is reached may be computed from equation qweb = 2q +FA1( 1y /I) q( just before level 1 or 2 ) = q (Flange A or B) + q ( of part of the web above level 1 or below level 2) = 7319.93+ FA y /I where A y =2.54*50.8(8.46 +76.2) = 13.109*103 mm3 q(just before level 1 or 2) = 7319.93+3336*13.109*103/11.196*10 6 =11.196*10 6 kN/m just at the level 1 or 2 , A y = area of Flange A or B * its y + area of flange C or D * its y + area of web above level 1 or below level 2 * its y = 2.54*76.2*127+2.54*50.8*76.2)*2.54*50.8*(25.4+76.2) = 47522 mm3

q ( level 1 or 2 ) = 3336*47522*10 9 / 11.196*10 6 = 14140.132 N/m q (level 1 or 2) = q (Flange A or B ) + q( Flange C or D) + shear Flow in the part of the web above level 1 or below level 2 = 7319.93 + 2933.2+3887.44 = 14140.57 N/m The maximum shear flow in the web occurs at the neutral axis where A y = area of cross section above or below the neutral axis multiplied by its y = 2.54*76.2*127+2.54*50.8*76.2 + 2.54*127*63.5 =53257.9 mm 3 q(neutral axis) = 3336*53257.9/11.196*10 6 = 16355 N/m q(neutral axis) = q(Flange A or B) + q(Flange C or D ) +Shear Flow in the web above or below the neutral axis = 7319.93+2933.2+ FA y /I =20483.8 mm 3 and FA y /I = 3336*20483.8*10 9 /11.196*10 6 = 6102.58 N/m


hence , q(neutral axis) = 7319.93+2933.2+6102.58 = 16355.27 N/m The shear flow for the entire section is as shown in Fig.(9-18 c ) Referring to Fig(9-18 d) , the shear center for the section is located from the relation

0Mo thus ,

1592.28*0.254 16.75*0.1524 F e = 0 putting F = P = 3336 N We obtain e = 17.85 mm , where e is the distance from the center line of the web to the shear center O. Alternatively the shear center O. may be located from equation

I4

tbhe22

Thus , for the flange A or B , h= 254mm , and b= 76.2mm and for the

Flange C or D , h = 152.4mm , and b =50.8mm

mm85.1710*196.11*4

54.2*8.50*4.15210*196.11*4

54.2*2.76*254e 6

22

6

22

We are able to show from the shear flow diagram of Fig(9-18c) that the applied load P = average shear flow in the web multiplied by the length of the web . The shear flow in the web shown in Fig(9-18c) may be drawn as shown in Fig(9-18e). Fig(9-18 f). Refer to Fig (9-18e) the average shear flow from the center line of the web to the vertex of the parabola shown dotted is q1(average) = 7319.9+(2/3) * height of parabola shown dotted = 7319.9 + (2/3) *6102.6 = 10.856 kN/m Refer to Fig(9-18 f) , the average shear flow from the vertex of the parabola shown dotted to the vertex of the parabola shown full is q2(average)=(14.14 13.422)+(16.355 14.14)*(2/3) + (13.422 11.207)*(1/3) = 2.933 kN/m Vertical shearing force = F = average shear flow in the web * length = q1(average) *0.254+q2(average)*0.1524 = 10856*0.254+2933*0.1524 = 3.336 kN = P


Example (9-12) Fig( 9 -19) shows a cantilever beam carries a downward vertical load P. If the cross section of the beam is a circular sector having the dimensions

shown in Fig(9-19 b ) locate the shear center O of the section . - a - - b Fig(9-19) Solution Refer to Fig. (9-19) , the moment of inertia of the section around the neutral axis of the section is determined as follows : Area of an element subtending an angle d is

dA = thickness * arc length = t. r d and the moment of inertia of the element about the neutral axis is dI = t. r d * square of the perpendicular distance from the centroid of


the element to the neutral axis

thus; dI = t. r d * r2 sin2(60 ) I = 2 t r3 3/

0

2 d)60(sin

But , 3/

0

2 d)60(sin = 3/

0

d)]2120cos(1[21

= 307.0)2120sin(21

21

3/

0

I = 0.614 t r3 ...................................... (9-12) Applying Eq.(9-8) , q = t a F y / I , and referring again to Fig (9-19b) , we have t . a = t . r d and

y = perpendicular distance from the centroid of the element to the neutral axis

= r * sin (60 )

hence , 0

)60sin(r rd.ty)a.t( ............... (9-13)

00

22 )]60[cos(r td)60sin(r ty)a.t(

= t r2 [cos (60 ) -14) The resultant force in the direction of the shear flow and perpendicular to the radius r is

dR = q * arc length = q * rd . see Fig(9-19a) the moment of this force about the center ( c ) of the circular sector is

dM = q r d (r) and total moment is M = 2 3

0

2 d qr .. . ( 9-1 5 )

Substituting q from Eq(9-8) and Eq(9-14) into Eq(9-15) , we obtain

M = 2 3/

0

3/

0

42 d)]2/1()60[cos(r t*)I/F(2dqr

= 2 ( F/I) r2 t [ sin (60 - 3/0]2

1) = 0.685 (F/I) r4 -16)

also the bending moment M about the center C of the circular sector is -17) From Eq(9-16) and Eq(9-17) we obtain .

P ( e + r) = 0.685 (F/I) t * r4 ................(9-18) Sub. For I from Eq(9-11) into Eq(9-18) and putting P = F , we obtain


r 116.0r0.614

r 685.0e


9-8 Problems 9-1) Fig(9-13) shows the web of a girder of I section is 45 cm deep and

1cm thick ; the Flanges are each 22.5 cm wide by 1.25 cm thick . The girder at some particular section , has to withstand a total shearing force of 200 kN . Find the shearing stresses at the top and middle of the web .

Fig(9-13) 9-2) Fig(9-14) shows a cast iron beam with an overhang . It has the

dimensioned channel section simply supported and subjected to the loads. If P1 = 0 , w = 17.5kN/m , and P2 = 0 , Find for the beam the position and magnitude of the max . shearing stress .

Fig(9-14) 9-3) Fig(9-15) shows a hollow circular section .Assuming the commonly

accepted theory of distribution of shearing stress in beams and starting from first principles , derive a formula for the intensity of shearing stress at any distance from the neutral axis in a hollow circular section , hence or otherwise show that in a very thin hollow rivet , the maximum shearing stress tends to be twice the mean shearing stress .


Fig(9-15) 9-4) An R.S.J. is of I section of overall height 200 mm and flange width

125 mm . The web thickness is 7 mm and the flange thickness 11mm. The standard taper on the flanges may be neglected and all corners may be assumed sharp. The beam is subjected to transverse loads acting parallel to the web , and at one section the shear force is 100 kN Find the max. vertical shearing stress in the web at this section. What proportion of the total shear force is carried by the web ?

9-5) Fig.(9-16) shows a section of steel bar is subjected to a shearing

force of 200 kN applied in the direction YY . Making the usual assumptions , find the average shearing stress at the sections A,B,C , and D , and find the ratio of the max. to the mean shearing stress in the section . Draw to scale , a diagram to show the variation of the average shearing stress across the section .

Fig(9-16)


9-6) Fig.(9-17) shows a circular ring , find the position of shear center in

a thin walled section in the form of a part of a circular ring with the central angle 2 , radius of central line r and constant thickness ( t ) .

Fig(9-17) 9-7) Fig.(9-18) shows a beam with its cross section. The beam is loaded

as shown in Fig. Draw diagrams of shearing stress , take MPa160

Fig(9-18) 9-8) Fig.(9-19) shows a beam with its cross section the beam is loaded as

shown in Fig. Find (i) the length of the beam (ii) max shearing stress take MPa 15.147 , MPa 24.39 ct

Fig(9-19) 9-9) Fig.(9-20) shows the cross-section of a beam . Find the distance ( e )

of the shear center from the center of gravity of the section ( consider the sections to be thin walled) .


Fig(9-20) 9-10) Fig.(9-21) shows the section of a horizontal cantilever which

carries a vertical end load W . Find the position of W relative to the edge AC in order that the cantilever shall not be subjected to torsion . ( work form first principle)

Fig(9-21)

Documents

Strength of Materials- Shear Stress in Beams- Hani Aziz Ameen