strength of materials- deflection of beams- hani aziz ameen

Strength of Materials Handout No.13

Deflection of 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.13- Deflection of Beams- Dr. Hani Aziz Ameen

13.1 Itroduction Under the action of applied forces the axis of a beam deflects from

its initial position. Basic differential equations for the deflection of beams will be developed in this chapter. 13 .2 Equation of the Elastic Curve

When a beam is loaded perpendicular to its axis ,as shown in Fig.(13-1) it defects from its initial position.

Fig(13-1) The deflection of the beam due to the applied loads should not be

excessive as it is always recommended that the max. deflection of the beam should not exceed 1/360 of its length. In order to derive the equation of the elastic curve consider any two points A and B on the curve as shown in Fig(13-2).

Fig(13-2) For convenience the arc AB whose length is ds is enlarged as shown in Fig(13-3). Let the length ds subtend an angle d , then ds = d

or dsdx

dxd

dsd1 (13-1)

Assuming that ds tends to zero ,

we have dxdy = tan

Fig(13-3)


dxd

dxdy1

dxdtan1

dxdsec

dxyd 2

222

2 .

(13-2) also, as ds tends to zero , (ds)2 = (dx)2 + (dy)2

or 22

dxdy1

dxds . (13-3)

substituting for dxd from Eq(13-2) and for

dxds from Eq(13-3) into

Eq(13-1) , yields

2

32

dxdy

2dx

y2d

1

1 (12-4)

Since the slope dxdy of the elastic curve is small ,

2

dxdy is still small

and therefore it is neglected, in case of beam deflected under a load , hence ,

2

2

dxyd1 .(13-5)

We established the equation M = EI in article (8-3) , substituting for

(1/R) from Eq.(13-5) gives

M = EI 2

2

dxyd .. (13-6)

It is known as the equation of the elastic or deflection curve The equation

EI 2

2

dxyd = M must be expressed in the form

E 2

2

dxyd =

IM , the function

IM then being integrated since F =

dxdM

and q = dxdF , from Article ( 7.3 , Eq.(8-4)) , it follows that EI 3

3

dxyd = F

and EI 4

4

dxyd = q


13.3 Methods of Computing the Deflection

The two methods that will be employed in computing the deflection of a loaded beam are the double integration method and the moment area method. 13.3.1 Double Integration Method

From equation M = EI 2

2

dxyd . If the bending moment can be expressed

as a function of x , successive integration will give expressions for

EI dxdy and EIy . The constants of integration may be determined from

the end fixing conditions.

EI 1)x( CdxMdxdy

EIy = 21)x( CxCdy dxM Where y is the deflection of the beam , and C1 & C2 are constant of integration.

13.3.1.1 Standard Case of Beam Deflections

(a) Cantilever Beam with Concentrated Load let y be the deflection at any distance x from the fixed end

M(x) = Ra . x+M = Wx +WL M(x) = W(L x)

EI 2

2

dxyd = W(L x)

EI dxdy = W(Lx

2x 2

) + C1

EIy = W (L6x

2x 32

) + C1x + C2

The Boundary condition

When x =0 , dxdy = 0 C1= 0

EIy = w ( 6

x2

Lx 32) + C2


when x = 0 , y = 0 C2 = 0

EIy = W ( 6

x2

Lx 32)

The max. slope and deflection occur at the free end, when x = L

i.e. EI2

WLdxdy 2

max

ymax = EI3

WL3

(b) Cantilever Beam with Uniform Distributed Load M(x) = Ra x + M+q x (x/2)

= q L.x + q 2Lq

2x 22

EI qdx

yd2

2(L x)

2xL

EI 2

2

dxyd (q/2) (L2 2Lx + x2)

EIdxdy = (q/2) (L2x 2L

3x

2x 32

) + C1

EIy = (q/2) (L2

12x

3xL

2x 432

) + C1x + C2


When x = 0 , 0dxdy , so that C1 = 0

EIy = (q/2) (12x

3xL

2xL

4322 ) + C2

When x = 0 , y = 0 , so that , C2 = 0 The max. slope and deflection occur at the free end . Where x = L

i.e. EI 6Lq

dxdy 3

max

ymax = EI8Lq 4


(c) Cantilever Beam with Moment at its End

M(x) = M

EI 2

2

dxyd = M

EI dxdy = Mx + C1

EIy = M 2

x2+ C1x + C2

The Boundary condition When x=0 , (dy/dx) = 0 , so that C1 = 0

EIy = M 2

x2+ C2

When x = 0 , y =0 C2 = 0 the max. slope and deflection occur at the free end , where x = L

EI

MLdxdy

max

ymax = EI2

ML2

(d) Cantilever Beam with Distributed Load

0Fy

Ra = 2Lqo

0MH

Ra L + 2Lqo

3L + M = 0

M = 2Lq 2

o 2Lqo

= 3Lq 2

o

M(x) = Ra .x M 3x

2x q

M(x) = 2Lqo x

3Lq 2

o L6xq 3

o

EI 2

2

dxyd =

2Lqo x

3Lq 2

o L6xq 3

o


EI

dxdy =

4Lqo x2

3Lq 2

o x L24

xq 4o + C1

EIy = 5x

L24q

2x

3Lq

3x

4Lq 5

o22

o3

o C1x + C2


At x = 0 , dxdy = 0 C1= 0

EIy = 3o x12

Lq 22

o x6Lq 5o x

L120q + C2

at x = 0 , y = 0 C2 =0

EIy = 3o x12

Lq 22

o x6Lq 5o x

L120q

The max. Slope and deflection occur at the free end where x = L

4o2

o2o

maxL

L24qL

3LqL

4Lq

EI1

dxdy

ymax. = 5o4

o3o LL120

q6LqL

12Lq

EI1


e) Simply Supported Beam with Central Concentrated Load Taking the original at the central

M(x) = Ra . ( x2L ) + Wx

= Wxx2W

2L

2W

EI 2

2

dxyd = x

2L

2W

EI 1

2C

2x

2Lx

2W

dxdy

EIy = 6

x4

Lx2W 32

+ C1x + C2


AT x = 0 , dxdy = 0 C1 = 0

When x = 2L , y = 0 C2 =

24L

2W 3

EIy = 24L

6x

4Lx

2W 332

The max. slope occurs at the ends, where x = 2L

EI16WL

dxdy 2

max

The max. deflection occurs at the center , where x = 0

EI48WLy

3

max


f ) Simply Supported Beam with Uniform Distributed Load

qLRR0F bay )2/L.(qLLR0M aA

Ra= qL /2 , R b= qL/2 0 x<L M(x) = Ra x q x (x/2) M(x)= (qL / 2) (qx2 / 2) EI (d2y/dx2) = (qLx/2) (qx2/2) EI (dy/dx) = (qLx2/4) (qx3/6)+C1 EIy = (qLx3/12) (qx4/24)+C1 x +C2 The Boundary conditions At x=0 , y=0 C2 =0 At x = L , y =0 C1 = ( qL3/24) (or we can apply other boundary condition is dy/dx = 0 , at x=L/2 ) EIy = ( qL/24) x4 + (qL/12)x3 (qL3/24) The max. deflection occurs at x = L/2 ymax = (( 5 q L4) / (384EI) ) 13.3.1.2 Macaulay s Method In this method we take whole the beam as the interval of x , and it is convential to use pointed brackets for terms such as < x a > , ( which is neglected in case of negative or zero ) to find M(x) Case (a) Concentrated Load 0 x<3a M(x) = Ra x P1 <x a > P2 < x 2a > EI (d2y/dx2) = M(x) EI (d2y/dx2) = Ra . x P1 < x a > P2 < x 2a >


EI (dy/dx) = Ra . x2/2 (P1 /2) < x a >2 (P2 /2) < x 2a >2 +C1 EIy = Ra .( x3/6 ) (P1 /6) < x a >3 (P2 /6) < x 2a >3 +C1 x + C2 Case ( b ) Applied Moment

M(x) = (M/L ) . x M < x a >0 , This term < x a >0 = 1 EI (d2y/dx2) = M(x) EI (d2y/dx2) = (Mx/L) M < x a>0 EI (dy/dx) = (M/L) (x2/2) M < x a >1 +C1 EIy = (M/6) x3 (M/2) < x a >2 + C1 x +C2 Case ( c ) Distribution Load If the beam shown in Fig (13-4a) carries a uniform distributed load q per unit length over the whole span the

M(x) = Ra. x W <x a> (q x2/2) If the distributed load only - a - covers the part DB (Fig.(13-4b)) M(x)=Ra.x W < x a > q < x b > <x b > / 2 M(x)=Ra.x W < x a > (q/2) < x b >2 - b - Fig(13-4)


If the distributed load does not continue to the end of the beam remote from the origin as shown in Fig(13-5)

Fig(13-5) It must be continue to the end and a compensating load added under neath as shown in Fig.(13-6)

Fig(13-6) M(x)=Ra x W < x a > q < x b > < x b > /2 + q < x c > < x c >/2 M(x) = Ra x W < x a > (q/2) < x b >2 + (q/2) <x c >2 In general , any distributed load , when started , must continue to the end remote from the origin , so that the load system shown in Fig.(13-7a) must be converted to that shown in Fig(13-7b) before Macaulay s Method can be applied .

Fig(13-7)


13.3.1.3 Examples The following examples explain the differences ideas of the deflection problems . Example (13-1) Fig.(13-8) show a cantilever beam . Find the ymax for the cantilever beam

Fig(13-8) Solution

PR0F ay a.PMMa.R0M aB

M(x)= Ra x M P < x a > EI (d2y/dx2) = P x P a P < x a > EI ( dy / dx) = P x2/2 P a x (P/2) < x a >2 + C1 (i) EIy = (P/6) x3 P.a (x2/2) (P/6) < x a >3 +C1 x +C2 (ii) Boundary conditions y = 0 at x = 0 , put in eq(ii) neglected 0 = (P/6) < a > + C2 C2 = 0 (dy/dx) = 0 at x=0 , put into Eq( i ) 0C1 y = (1/EI) ( (Px3/6) (Pa/2) x2 (P/6) < x a >3 ) ymax at x = L ymax = (1/EI) *( (PL3/6) (P a L2 /2) (P/6) < L a >3) = (PL3/6) (PaL2/2) (P/6) [(L a)(L2 2aL+a2)] = (PL3/6) (PaL2/2) (P/6) [L3 2aL+a2L aL2+2a2L a3]


= Pa2L( 3

161 )+ Pa3/6 = Pa2L ( 6

21 ) + (Pa3/6)

ymax = (1/EI) (( Pa2/6) (3L a))

Example (13-2) Fig.(13-9) shows the simply supported beam 5 m long, subjected to the uniform distributed load at 3m from the left side as shown in the figure Find the deflection at the right end .

Fig(13-9) Solution M(x) = 650 x 400 (x2/2) + (400/2) < x 3 >2 + 950 < x 4 > EI (dy/dx) = 650 (x2/2) 400 (x3/6) + (400/60) < x 3 >3 +(950/2) < x 4 > 2 + C1 EIy = 650 (x3/6) 400 (x4/24) + (400/24) < x 3 >4 +(950/6) < x 4 > 3 + C1x+C2 Boundary condition y = 0 at x = 0 C2 = 0 y = 0 at x = 4 C1= 671 y = (1/EI) * [(325x3/3) (50x4/4) + (50/3) < x 3>4 +(475/3) <x 4 >3 671x ] at right hand x = 5 y = (1/EI) * 195


Example(13-3) Fig.(13-10) shows a simply supported beam subjected to the load indicated in the figure. Find the midspan deflection , select origin at midspan position of the elastic curve .

Fig(13-10) Solution

Vo=(wL/4) (wL/4) = 0 Mo = (wL/4)(L/2) (wL/2)(1/2) [(L/2 ).(2/3)] = (1/24) wL2 EI (d2y/dx2) = (wL2/24) (wx3/3L) EI(dy/dx) = (wL2/24)x (w/12L)x4 + C1 EIy = (wL2/48) x2 (w/60L)x5 + C1 x +C2 Boundary Condition y = 0 at x = 0 C2 = 0 dy/dx = 0 at x=0 C1=0 y= (1/EI) [ ( wL2/48)x2 (w/60L) x5] ymax at x = L/2 ymax = (9 w L4 )/ (1920 EI)


Example(13-4) Fig.(13-11) shows a beam subjected to a concentrated load (P) . Find ymax .

Fig(13-11) Solution M(x) = Ra . x P < x a > EI(d2y/dx2) Ra x P < x a > EI ( dy/dx) = Ra (x2/2) (P/2) < x a >2 + C1 EIy = (P.b/L)(x3/6) (P/6) < x a>3 + C1x + C2 .. (i) The boundary condition x=0 at y=0 C2 = 0 x=L at y = 0 0 = (P.b/L) (L3/6) (P/6) < L a>3 +C1L

L

)aaL2L)(aL)(6/P()6/L)(L/Pb(C223

1

at maximum deflection , the slope is horizontal i.e. dy/dx = 0 0 = Ra (x2/2) P <x a>2 + C1 Ra (x2/2) (P/2) ( x2 2ax + a2) +C1 (P.b/4)x2 (P/2) x2 + (P/2) 2 a x (P/2)a2 +C1 = 0

we obtain the value of 3

aLx22

sub into Eq(i) , the max. deflection is

LEI 39)aL.(a.Py

2/322

max


Example(13-5) Fig.(13-12) shows a wooden flag-post 6 m high , is 50 mm square for the upper 3m and 100mm square for the lower 3 m . Find the deflection of the top due to a horizontal pull of 40 N at that point applied in a direction parallel to one edge of the section E = 10 GPa .

Fig(13-12) Solution The total defection at the top is made up of:

(1) The deflection at B (y1). (2) The slope at B, multiplied by the distance BC (y2). (3) The further deflection due to bending of BC (y3).

These deflection are shown in Fig(13-12c). Let the second moments of area of parts AB and BC be I1 & I2 receptively. Then y=y1+y2+y3

2

3

11

2

1

2

1

3

EI33*403*

EI3*120

EI33*40

EI23*120

EI33*40

m09936.005.01

1.07

10*1012*360

449


Example(13-6) Fig.(13-13) shows the two beams AB and CD. are of the same material and have the same cross-section. The support at B is at the same level as the fixed end A. Find the reactions at B and D if the beam CD arrives on

its whole length a uniformly distributed load of 1 kN/m. Fig(13-13) Solution Let reaction at B be R and the force in the spacer at D be P, as shown below

Then the downward deflection at B due P must equal the upward deflection due to R, since the point B is level with A,

i.e. EI3

5*R1*EI24*P

EI34*P 323

R88

125P ( i )

The deflection at D must be the same for the upper beam as it is for the lower beam, since the spacer is assumed rigid. For the deflection at D on the lower beam, it will be convenient to move the force R to D and introduce an anticlockwise moment R*1 to compensate.

i.e. EI24*R

EI3RP

EI34*P

EI84*1 234

from which 16P 11R=12 ( ii ) from Eq(i) & Eq(ii) , P=1.454 kN

R=1.023 kN


Example(13-7) Fig(13-14) shows the overhung crankpin of a locomotive can be considered as a cantilever of length L , and the distributed load applied to the pin by the hydrodynamic lubricating film can be assumed to be of the form k(Lx-x2) per unit length x is the distance from the built in end and k is a constant. Find the expression for the deflection at the free end of the pin.

Fig(13-14) Solution

24

4xLxk

dxydEI

1

32

3

3C

3x

2Lxk

dxydEI

When x=L, S.F.=0, so that C1= 6kL3

2

343

2

2C

6xL

12x

6Lxk

dxydEI

When x=L, B.M.=0, so that C2= 12kL4

3

42354C

12xL

12xL

60x

24Lxk

dxdyEI

When x=0, 0dxdy , so that C3=0

4

243365C

24xL

36xL

360x

120LxkEIy

When x = 0 , y = 0, so that C4 = 0 Therefore the deflection at free end

EI360kL7

24L

36L

360L

120L

EIk 66666


Example(13-8) Fig.(13-15) shows a cantilever of circular section tapers uniformly from a diameter D at the free end to 2D at the fixed end. It carries a single concentrated load at the free end. Find the diameter of a cantilever of uniform diameter, which would have the same end deflection. Prove any formula used for calculating the deflection of the tapered cantilever.

Fig(13-15) Solution The simplest expression for the second moment of area of a typical section will be obtained by taking the point O as the origin for x.

Then diameter of section = DLx

44

4D

Lx

64I

43

44

42

2Lxxk

DLx

64

LxWI

Mdx

ydE

where 4

4

DWL64k

1

32C

3Lx

2xk

dxdyE

when x=2L, 0dxdy , so that C1= 2L12

k

Ey = k [ ( x 1 /2 ( Lx 2 )/6+ x /(12L2) ] +C2 When x = 2 L , y=0 C2 = 3k/(8L) When x = L , y= (k/E) [ (1/(2L)) (1/(6L2))+(1/(12L2) (3/(8L))] y= (64WL3) / (24E D4)

Deflection at free end of cantilever of uniform diameter = 464

3

dE3WL

4

3

4

3

dE3WL64

DE24WL64


d= D 4 8 = 1.682 D Example(13-9) Fig(13-16) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy .

Fig(13-16) Solution R1+R2 = 12 4*5 R2 + 4*2 +4 =0

R2 = 8 kN R1 = 4 kN

M(x) = R1x 2 (x2/2) 4 <x 2> + (2/2) < x 2>2+R2<x 4> (2/2)<x 4>2 M(x)= 4x x2 <x 2> + <x 2>2+ R2 <x 4> <x 4>2 EI(dy/dx) =4x2/2 x3 /3 (4/2) <x 2>2 + (1/3)<x 2>3 + (8/2)<x 4>2 (1/3) < x 4 >3 +C1 EIy =2x3/3 (1/3)(x4 /4) (2/3) <x 2>3 + (1/3)(1/4)<x 2>4

+ (4/3)<x 4> (1/3) < x 4 >4 +C1x+C2 EIy = (2/3)x3 x4/12 (2/3) <x 2>3 + (1/12)<x 2>4+(4/3)<x 4> (1/3) <x 4>4 +C1x+C2 The boundary conditions y = 0 at x = 0 0C2 and y = 0 at x = 4 0 = (2/3)(4)3 (44/12) (2/3)<4 2>3 + (1/12) < x 2 >4 + C1*4 C1 = 4.33 EIy = (2/3)(2)3 (1/3)(24/4) (2/3)<2 2>3+(1/12)<2 2>4+(3/4)<2 4 > (1/3) <2 4 >4 + ( 4.33)*2 EIy = 4.66 kN.m3


Example(13-10) Fig(13-17 ) shows the simply supported beam subjected to the loads

indicated in the figure . Find the value of EIy . Fig(13-17) Solution R1+R2=600 4R1 600*1.5=0 R1=225 N R2=375 N 4R2+600*2.5=0 R2=375 N R1=225 N M(x) =R1x (300/2) <x 1.5>2 + (300/2)<x 3.5>2

EIxdyd

2

2=255x 150<x 1.5>2+150<x 3.5>2

EIdxdy = 1

332 C5.3x3

1505.1x3

150x2

225

EIy= 21443 CxC5.3x

4505.1x

450x

35.112

y = 0 at x = 0 C2 = 0 y = 0 at x = 4 m

1443 C45.34

4505.14

450)4(

35.1120

C1= 478.125

EIy = 2*)125.478(5.324

505.144

50)2(3

5.112 443

EIy = 657 N.m3


Example(13-11) Fig(13-18) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy .

Fig(13-18) Solution R1+ R2 = 1200 + 400 = 1600 4R1 1200*2.5+400=0 R1=650 N R2=950 N M(x) =R1.x 400(x2/2)+R2 <x 4> + (400/2) < x 3 >2

EIxdyd

2

2=650x 200x2+950<x 4>+200<x 3>2

EIdxdy = 1

3232 C3x3

2004x2

950x3

200x2

650

EIy = 214443 CxC3x

47.664x

3475x

47.66x

3325

y= 0 at x = 0 C2 = 0 y = 0 at x = 4

14443 C434

47.6644

3475)4(

47.66)4(

33250

C1= 670.3

EIy = )3.670(*5354

7.66453

475)5(4

7.66)5(3

325 4443

EIy = 193.42 N.m3


Example(13-12) Fig(13-19) shows the simply supported beam subjected to the loads indicated in the figure . Find the value of EIy .

Fig(13-19) Solution R1+R2=1600 6R1 800*5 800=0 R1=800 N R2=800 N M(x) =R1.x 400(x2/2)+(400/2) <x 2>2+(400/2)<x 4>2

EIxdyd

2

2=800x 200x2 + 200 <x 2>2 + 200 <x 4> 2

EIdxdy = 1

3332

C4x3

2002x3

2003

x2002

x800

EIy = 214443 CxC4x

47.662x

3475x

47.66x

3400

y = 0 at x = 0 C2 = 0 y = 0 at x = 6

1443 C646

47.6626

47.66)6(

47.66)6(

34000

C1= 1865.2

EIy = 3*2.1865434

7.66234

7.66)3(4

7.66)3(3

400 4443

= 3.3296 N.m3


13-3-2 Moment Area Method

The alternative method of computing the deflection of beams is the moment area method.

The deflection computed by the moment area method can be considerably simplified especially if the deflection is required at a specified point on the beam.

Fig(13-20) Consider any two points C & D on the elastic cure of the beam, as

shown in Fig(13-20b) which is loaded arbitrarily as shown in Fig(13-20b) if tangents are taken at points A & B, they will intersect at an angle d . The arc length of CD = ds The deflection of any point on the beam is so small. Hence the length ds = dx ds = d

dxd

dsd1

i.e. hence EIM1 (from article 8-3 )


dx

EIMd -7)

Form the above derivation it can be concluded :- First theorem

curve is

The deviation dt = x d

B

A

B

AA/B xddtt -8)

Substituting Eq(13-7) into Eq(13-8) gives

B

AA/B dx

EIMxt

XB

XAA/B Mdxx

EI1t

where (M dx)= shaded area (Fig(13-20c)) tB/A (M diagram / EI) multiplied by the horizontal distance from the center of gravity of the whole shaded area to the vertical through point B. Second theorem

equal to EI1

BABA/B x.areaEI1t

where Bx is the horizontal distance from the center of gravity of the whole shaded area to the vertical through point B. (see Fig(13-20c)). Note: The deviations is very often different from the deflection.

strength of materials- deflection of beams- hani aziz ameen

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