Material of Strainght II 2011

8/4/2019 Material of Strainght II 2011

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Chapter 2:Deflection of beamIntroduction

Why we have to design of beam ?

a) Find the limitation of yield stressb) Measure maximum value of

deflection of beam when have load


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Purposes ?

a) Shaft with gear or component

Gear teeth not fit

Friction , vibration

Noisy , failure

b) Floor of building

Plaster ceiling will fracture

c) Structure of an aeroplane Vibration

Not balance


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A beam is a member subjected to loads applied

transverse to the long dimension, causing the member to

bend. For example, a simply-supported beam loaded at itsthird-points will deform into the exaggerated bent shape

shown in Fig. 2.1

Figure 2.1 Example of a bent beam (loaded at its third points)


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Beams are frequently classified on the basis of supports or

reactions.

A simple beam :

i. beam supported by pins,

ii. rollers, or

iii. smooth surfaces at the ends A simple support develop a reaction normal to the beam, not

produce a moment at the reaction.

Both ends of a beam projects beyond the supports, it is called

a simple beam with overhang. A beam with more than simple supports is a continuous beam.


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Figure 2.2 Various types of beams and their deflected shapes:

a) simple beam,

b) Beam with overhang,

c) continuous beam


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Figure 2.2 Various types of beams and their deflected shapes:

d) a cantilever beam,

e) a beam fixed (or restrained)at the left end and simply supported near the

other end (which has an overhang),

f) beam fixed (or restrained) at both ends.


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Deflections Due to Moments: When a straight beam is loaded and the

action is elastic, the longitudinal centroid axis of the beam becomes a

curve defined as "elastic curve." In regions of constant bending moment,

the elastic curve is an arc of a circle of radius, r, as shown in Fig. 2.3 in

which the portion AB of a beam is bent only with bending moments.

Therefore, the plane sections A and B remain plane and the deformation

(elongation and compression) of the fibres is proportional to the distance

from the neutral surface, which is unchanged in length. From Fig. 2.3:

Figure 2.3 Bent element from which relation for elastic curve is obtained


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which relates the radius of curvature of the neutral surface of the beam

to the bending moment, M, the stiffness of the material, E, and the

moment of inertia of the cross section,I.


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Important equation

EI d2y = EIy = Mx-x . momentdx

EI dy = EIy = Mdx + A slope ( rad)dx

EIy = [ Mdx ]dx + Ax + B .. Deflection (mm @m)


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Example 1:

Figure below shown a simply supported beam at point A and B with load at

the middle of the beam:

a) Maximum of slope

b) Maximum of deflection

A B

W

L


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Solution 1:

Free body diagram

RA

L/2

RB

L/2

W

+ ve MA = 0

W(L/2)RB (L) = 0

RB (L) = W(L/2)

RB = W/2

+ ve Fy = 0

RA W + RB = 0

RA = W RB

RA = W W/2RA = W/2


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Section x-x from point A

+ve Mx-x = RA(x) = Wx/2

RA

L/2

RB

L/2

W

x

x

x

EI d2y = EIy = Mx-x = Wx/2 momentdx

EI dy = EIy = Mdx + A = W(x/2) dx slopedx

= Wx2 / 4 + A

EIy = [ Mdx ]dx + Ax + B = [Wx2 / 4 + A ]dx .. Deflection

= Wx3 / 12 + Ax + B


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Find the value of A and B

When x = L/2 , y = 0

EIy = Wx2 / 4 + A slope

0 = W (L/2)2 / 4 + A

W (L2 / 4 ) / 4 = -A

W (L2 / 16 ) = - A

A = - W L2 / 16

replace in slope eq.

EIy = Wx2 / 4 + A slope

= Wx2 / 4 - W L2 / 16


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When x = 0 , y = 0

EIy = [Wx2 / 4 + A ]dx .. Deflection

= Wx3 / 12 + Ax + B

0 = 0 + ( - W L2 / 16)x + B

0 = 0 + 0 + B

B = 0replace in deflection eq.

EIy = Wx3 / 12 + ( - W L2 / 16)x .. Deflection


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a) Maximum of slope

occur when x = 0 and x = L means at point A and point B

slope eq.

EIy = Wx2 / 4 - W L2 / 16

At point A , x = 0

EIy = Wx2 / 4 - W L2 / 16

EIyA = W(0)2 / 4 - W L2 / 16EIyA = - W L2 / 16

yA = - W L2 / 16EI rad

At point B , x = L

EIy = Wx2 / 4 - W L2 / 16

EIyB = WL2 / 4 - W L2 / 16

EIyB = (4WL2 - W L2 )/ 16

yB =

3W L

2

/ 16EI rad


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b) Maximum of deflection

when x = L/2

replace in deflection eq.

EIy = Wx3 / 12 + ( - W L2 / 16)x .. Deflection

When x = L/2

EIy(x=L/2) = W(L/2)3

/ 12 + ( - W L2

/ 16)(L/2)

EIy = WL3 /96 - W L3 / 32

EIy = (WL3 - 3 W L3 )/ 96

EIy = - 2WL3 / 96

y = - WL3 / 48EI (mm @ m)


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Example 2:

Find the slope and deflection at point A when the cantilever beam AB

carries a load(KN) and uniformly distributed load W (KN/m). Assume EI (KNm2

)as a constant for beam. Then measure the new slope and deflection at the

same point when:

a) W = 0

b) P = 0

AB

W (KN/m)

L

P


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+ve Mx-x =PxWx2

2

EI d2y = EIy = Mx-x = Px Wx2 . moment (1)dx 2

EI dy = EIy = Mdx + A =Px2Wx3 + A slope (2)dx 2 6

EIy = [ Mdx ]dx + Ax + B = Px3Wx4 + Ax + B Deflection (3)6 24

Solution 2:


Free body diagram

A

FB

W (KN/m)

L

Px

x

x

MB


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Refer to boundary condition :

a) When ( x = L , y = 0 )

b) When ( x = L , y = 0 )


AB

W (KN/m)

L

Px

x

x


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a)When ( x = L , y = 0 )


EI dy = EIy = Mdx + A =Px2Wx3 + A slope (2)dx 2 6

EIy = Px2Wx3 + A2 6

0 =P(L)2W(L)3 + A2 6

A = PL2 + WL3

2 6


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When ( x = L, y = 0 ) And A = PL2 + WL3

2 6


EIy = [ Mdx ]dx + Ax + B = Px3Wx4 + Ax + B Deflection (3)6 24

EIy = Px3Wx4 + Ax + B

6 240 = P(L)3W(L)4 + A(L) + B

6 240 = PL3WL4 +( PL2 + WL3 )L + B

6 24 2 60 = PL3WL4 + PL3 + WL4 + B

6 24 2 60 = PL3 + 3PL3 WL4 + 4WL4 + B

6 6 24 240 = 2PL3 + 3WL4 + B

6 24B = [ PL3 + WL4 ]

3 8


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EIy = Px2Wx3 + PL2 + WL3 slope (2)2 6 2 6

EIy = Px3Wx4 + ( PL2 + WL3 )x [ PL3 + WL4 ] Deflection (3)6 24 2 6 3 8

EIy = Px3Wx4 + PL2 x + WL3x PL3 WL4 Deflection (3)6 24 2 6 3 8

New slope and deflection equations:

Q : Find the slope and deflection at point A

When x = 0 (y = ????)

EIy = Px2Wx3 + PL2 + WL3

2 6 2 6EIy = P(0)2W(0)3 + PL2 + WL3

2 6 2 6EIy = 00 + PL2 + WL3

2 6yA = 1 ( PL

2 + WL3 ) rad

EI 2 6


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EIy = Px3Wx4 + PL2 x + WL3x PL3 WL4 Deflection (3)6 24 2 6 3 8

New slope and deflection equations:

Q : Find the slope and deflection at point A

When x = 0 ( y = ???? )

EIy = Px3Wx4 + PL2x + WL 3x PL3 WL4

6 24 2 6 3 8

EIy = P(0)3W(0)4 + PL2 (0) + WL3 (0) PL3 WL4

6 24 2 6 3 8

EIy = P(0)3W(0)4 + PL2 (0) + WL3 (0) PL3 WL4

6 24 2 6 3 8

EIy = 00 + 0 + 0 PL3 WL4

3 8

yA = 1 ( PL3 WL4 ) mm@ m

EI 3 8


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Q : Find the slope and deflection at point A when :

a) W = 0

yA = 1 ( PL2 + WL3 ) radEI 2 6

yA = 1 ( PL2 + (0)L3 )EI 2 6

yA = PL2 rad

2EI

yA = 1 ( PL3 WL4 ) mm@ mEI 3 8

yA = 1 ( PL3 (0)L4 )

EI 3 8yA = 1 ( PL3 )

EI 3yA = PL

3 mm@ m3EI


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Q : Find the slope and deflection at point A when :

b) P = 0

yA = 1 ( PL2 + WL3 ) radEI 2 6

yA = 1 ( (0)L2 + WL3 )EI 2 6

yA = WL3 rad

6EI

yA = 1 ( PL3 WL4 ) mm@ mEI 3 8

yA = 1 ( (0)L3 WL4 ) mm@ m

EI 3 8yA = WL

4 mm@ m8EI


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Example 3:

The cantilever beam shown is subjected to a vertical load 7KN at B

point. Give the cross section of the beam is 100mm x 300mm for

length and depth and the materials modulus of elasticity is 8GPa.

Determine :

a) The slope and deflection at point B.

b) The deflection at the end of cantilever beam.

A

B

7 KN

2m

C

1.5m


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Solution 3:

Free body diagram

FAB

7 KN

2m

C

1.5m

MA

+ ve M = 0

MA + 7(2)= 0

MA = -14KNm

+ ve Fy = 0

FA 7 = 0

FA = 7KN


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EI d2y = EIy = Mx-x = -14 + 7x 7(x 2) . moment (1)dx

EI dy = EIy = Mdx + A = -14x + 7x2 7(x2)2 + A slope (2)dx 2 2

EIy = [ Mdx ]dx + Ax + B = -14x2 + 7x3 7(x2)3 + Ax + B Deflection (3)2 6 6

Solution 3:

Free body diagram


+ve Mx-x = MA + FA(x) 7(x2)

Mx-x = -14 + 7x 7(x2)

FAB

7 KN

2m

C

x

MA

x

x


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a) When ( x = 0 , y = 0 )

b) When ( x = 0 , y = 0 )


FAB

7 KN

2m

C

x

MA

x

x



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EIy = -14x + 7x2 7(x2)2 + A slope (2)2 2

EIy = -14x2 + 7x3 7(x2)3 + Ax + B Deflection (3)2 6 6

EIy = -14x + 7x2 7(x2)2 + A2 2

0 = - 14(0) + 7(0)2 7((02)2 + A2 2

0 = 0 + 00 + A

A = 0

a) When ( x = 0 , y = 0 ) replace in deflection eq.

b) When ( x = 0 , y = 0 ) replace in slope eq.

EIy = -14x2 + 7x3 7(x2)3 + Ax + B2 6 6

0 = -14(0)2

+ 7(0)3

7(02)3

+ 0 + B2 6 6

0 = 0 + 0 0 + 0 + B

B = 0


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EIy = -14x + 7x2 7(x2)2 new slope eq. (2)2 2

EIy = -14x2 + 7x3 7(x2)3 new deflection eq. (3)2 6 6

I = bd3 = (0.1)(0.3)3 = 225 x 10-6m4

12 12

E = 8GPa

Q: Determine

a) The slope and deflection at point B.

b) The deflection at the end of cantilever beam.


a) when x = 2 , yB = ?? replace in deflection eq.

when x = 2 , yB = ?? replace in slope eq.

b) when x = 3.5 , yC = ?? replace in new deflection eq.


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a) when x = 2 , yB = ?? replace in deflection eq.

EIy = -14x2 + 7x3 7(x2)3 new deflection eq. (3)

2 6 6

EIyB = -14(2)2 + 7(2)3 7(22)3

2 6 6

EIyB = -28 + 56 06

yB = -168 + 56 = _____-112 x 103

6EI 6 x 8 x109 x 225 x 10-6

yB = -10.37mm


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a) when x = 2 , yB = ?? replace in slope eq.

EIy = -14x + 7x2 7(x2)2 new slope eq. (2)

2 2

EIyB = -14(2) + 7(2)2 7(22)2

2 2

EIyB = -28 + 14

yB = -14 = -14 x 103

EI 8 x109 x 225 x 10-6

yB = - 7.78 x 10-3 rad


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EIy = 14x + 7x2 7(x2)2 new slope eq. (2)2 2

EIy =14x2 + 7x3 7(x2)3 new deflection eq. (3)2 6 6


b) when x = 3.5 , yC = ?? replace in new deflection eq.

EIy =14x2 + 7x3 7(x2)3 new deflection eq. (3)

2 6 6

EIyC =14(3.5)2 + 7(3.5)3 7(3.52)3

2 6 6

EIyC = 85.75 + 50.02 3.94

yC = -39.67 = - 39.67 x 103

EI 8 x109 x 225 x 10-6

yC = -22.04mm


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Macaulay FunctionExample of cases:

W(KN/m)

W(KN/m)

W(KN/m)

E l 4


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Example 4:

Figure below shown a simply supported beam at point A and B

with load 20KN at the end. Given EI = 5.0KNm2

a) Deflection equation by using Macaulay Method

b) Deflection at 20KN point

20KN/m

20KN

BA

2m 2m 1m

S l ti 4


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+ ve MB = 0

20(2)(3) + RA (4) + 20(1) = 0

4RA = 12020

RA = 25KN

+ ve Fy = 0

RA 20(2) + RB 20 = 0

RB = 40 +20 25

RB = 35KN

Solution 4:

Free body diagram

20KN/m

20KN

BA

2m 2m 1m

S ti f i t A


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20KN/m

20KN

35KN25KN2m 2m 1m

xx

x

+ve Mx-x = 25x20x2 + 20(x2)2 + 35(x4)2 2

EI d2y = EIy = Mx-x =25x20x2 + 20(x2)2 + 35(x 4) . moment (1)dx 2 2

EI dy = EIy = 25x220x3 + 20(x2)3 + 35(x4)2 + A slope (2)dx 2 6 6 2

EIy = 25x320x4 + 20(x2)4 + 35(x4)3 + Ax + B Deflection (3)6 24 24 6


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replace in deflection eq. or eq. (3)EIy = 25x320x4 + 20(x2)4 + 35(x4)3 + Ax + B Deflection (3)

6 24 24 6


a) When ( x = 0 , y = 0 )b) When ( x = 4 , y = 0 )



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replace in deflection eq. or eq. (3)

EIy = 25x320x4 + 20(x2)4 + 35(x4)3 + Ax + B Deflection (3)6 24 24 6

a) When ( x = 0 , y = 0 )

EIy = 25x320x4 + 20(x2)4 + 35(x4)3 + Ax + B6 24 24 6

0 = 25(0)3

20(0)4

+ 20(02)4

+ 35(04)3

+ A(0) + B6 24 24 6B = 0

b) When ( x = 4 , y = 0 )EIy = 25x320x4 + 20(x2)4 + 35(x4)3 + Ax + B

6 24 24 60 = 25(4)320(4)4 + 20(42)4 + 35(44)3 + A(4) + (0)

6 24 24 60 = 266.67213.33 + 13.33 +4AA = -16.67


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EIy = 25x320x4 + 20(x2)4 + 35(x4)3 16.67x6 24 24 6

a) Deflection equation by using Macaulay Method


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replace in deflection eq. or eq. (3)

EIy = 25x320x4 + 20(x2)4 + 35(x4)3 16.67x Deflection (3)6 24 24 6

b) Deflection at 20KN point

When x = 5 , y20 = ????

EIy = 25x320x4 + 20(x2)4 + 35(x4)316.67x6 24 24 6

EIy20 = 25(5)320(5)4 + 20(52)4 + 35(54)3

16.67(5)6 24 24 6

EIy20 = 520.8520.8 + 67.5 + 5.883.35

y20 = 10.08 x 103 =10.08 x 103

EI 5 x 103

y =2.016m

Material of Strainght II 2011

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