JJ310 STRENGTH OF MATERIAL Chapter 5(a) Beam Deflection

CHAPTER 5 (a)

BEAM DEFLECTION

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Learning Outcomes:At the end of this lecture, student should be able to;

Understand beam deflectionExplain double integration methodState the boundary condition for simply supported beamDetermine the slope and deflection in every loaded situation as below:Simply support, with concentrated load at center of the beam.Simply support, with uniform distributed load.

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Beam Deflection Beam deflection is defined as the distance from the neutral plane origin to the position of neutral plane when the load is applied.

ymaxF = Load y = DeflectionNeutral axis originNeutral axis after load applied*

General Formula M = EI d2y Moment Equation dx2

Where;M = Moment (Nm) I = Second moment of area @ moment inertia (m4)E = Young Modulus (Pa)

* EI also known as flexural rigidity

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Double Integration Method

EI d2y = M (Moment Equation) dx2 If the beam has uniform cross-section and M can be expressedmathematically as x function, therefore;

EI dy = Mdx + A (Slope Equation) dx

EIy = Mdx + Ax + B (Deflection Equation)

Notes: dy/dx = Slope (no unit) y = Deflection (m or mm)

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Case 1: Simply supported beam, with concentrated load at center (boundary conditions included)

Consider this simply supported beam AB, with length L and carry the concentrated load at center. Take A as the reference point and by balancing the moment and force we get;

RA = RB = F/2

On the x-x section (distance x is based on point A), it is known that the moment at this section is;M xx = Fx / 2ABx = Ldy/dx = maxy = 0x = 0dy/dx = maxy = 0 x = L/2dy/dx = 0y = max

xx*

General Formula

a) EI d2y = M (Moment Eq.) dx2

b) EI dy = Mdx + A (Slope Eq.) dx

c) EIy = Mdx + Ax + B (Deflection Eq.)

Solution

Substitute M = Fx / 2 into eq. (a)

EI d2y = Fx / 2 dx2

Integrate once we get;

EI dy = Fx2 + A dx 4

Integrate once again we get;

EIy = Fx3 + Ax + B 12

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Boundary Conditionsa) To find the variables A and BEI dy = Fx2 + A .. (Slope Eq.) dx 4

When; x = L/2 , dy/dx = 0

Thus;EI (0) = F(L/2)2 + A 4 A = - FL2 16*

EIy = Fx3 + Ax + B (Deflection Eq.) 12

Substitute A = - FL2 16

EIy = Fx3 FL2x + B 12 16

When; x = 0, y = 0

Thus;EI (0) = F(0)3 FL2(0) + B 12 16 B = 0

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Therefore, the COMPLETE slope and deflection equations are;

i) EI dy = Fx2 - FL2 . (Slope Eq.) dx 4 16

ii) EIy = Fx3 FL2x ..... (Deflection Eq.) 12 16

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b) To find the Maximum slope and deflection

i)Maximum Slope

EI dy = Fx2 - FL2 dx 4 16

When; x = 0,x = L

EI dy = F(0)2 - FL2 dx 4 16

dy = - FL2dx16EI

-ve sign indicates that the deflections is below the undeformed neutral axis *

ii) Maximum deflection

EIy = Fx3 FL2x 12 16

When;x = L/2

EIy = F(L/2)3 FL2 (L/2) 12 16

EIy = - FL3 48

y = - FL3 48EI

-ve sign indicates that the deflections is below the undeformed neutral axis *

Case 2: Simply supported beam, with uniformly distributed load (boundary conditions included)

Consider this simply supported beam AB, with length L and carry the uniformly distributed load along the beam. Take A as the reference point and by balancing the moment and force we get;

RA = RB = FL/2

On the x-x section (distance x is based on point A), it is known that the moment at this section is;M xx = wLx / 2 Wx2/2ABx = Ldy/dx = maxy = 0x = 0dy/dx = maxy = 0 x = L/2dy/dx = 0y = max

xx*

General Formula

a) EI d2y = M (Moment Eq.) dx2

b) EI dy = Mdx + A (Slope Eq.) dx

c) EIy = Mdx + Ax + B (Deflection Eq.)

Solution

Substitute M = wLx wx2 into eq. (a)2 2

EI d2y = wLx wx2 dx2 2 2

Integrate once we get;

EI dy = wLx2 - wx3 + A dx 4 6

Integrate once again we get;

EIy = wLx3 wx4 + Ax + B 12 24

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Boundary Conditionsa) To find the variables A and BEI dy = wLx2 - wx3 + A .. (Slope Eq.) dx 4 6

When; x = L/2 , dy/dx = 0

Thus;EI (0) = wL(L/2)2 w(L/2)3 + A 4 6 A = - wL3 24*

EIy = wLx3 wx4 + Ax + B (Deflection Eq.) 12 24

Substitute A = - wL3 24

EIy = wLx3 wx4 wL3x + B 12 24 24

When; x = 0, y = 0

Thus;EI (0) = wL(0)3 w(0)4 wL3(0) + B 12 24 24 B = 0

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Therefore, the COMPLETE slope and deflection equations are;

i) EI dy = wLx2 - wx3 wL3 . (Slope Eq.) dx 4 6 24

ii) EIy = wLx3 wx4 wL3x ..... (Deflection Eq.) 12 24 24

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b) To find the Maximum slope and deflection

i)Maximum Slope

EI dy = wLx2 - wx3 - wL3 dx 4 6 24

When x = 0,EI dy = wL(0)2 - w(0)3 wL3 dx 4 6 24

dy = - wL3 dx 24EI

When x = LEI dy = wL(L)2 - w(L)3 wL3 dx 4 6 24

dy = + wL3 dx 24EI

+ve sign indicates that the deflections is above the undeformed neutral axis *

ii) Maximum deflection

EIy = wLx3 wx4 wL3x 12 24 24

When;x = L/2

EIy = wL(L/2)3 w(L/2)4 - wL3(L/2) 12 24 24

EIy = - 5wL4 384

y = - 5wL4 384EI

-ve sign indicates that the deflections is below the undeformed neutral axis *

Exercise1. A simply supported beam is 4m long and has a load of 200KN at the middle. The flexural stiffness is 300MPa. Calculate the slope at the ends and the deflection at the middle. (Answer: 6.67 x 10-4 , 0.89 mm)

2. A simply supported beam is 8m long with a load of 500KN at the middle. The deflection at the middle is 2 mm downwards. Calculate the slope at the ends. (Answer : 750 x 10-6)

3. A simply supported beam is 4m long with a UDL of 4KN/m. The dimension of the beam is 300mm x 100mm. Calculate the maximum slope and deflection. Given E = 200 GPa. (Answer: 2.13 x 10-3 , 2.7 x 10-3m)

4. A simply supported beam is 5m long with UDL of 200 N/m. The flexural stiffness is 100MPa. Calculate the slope at the ends and the deflection at middle. (Answer: 1.04 x 10-5, 1.63 x 10-5 m*