CHAPTER 5 (b)

BEAM DEFLECTION*

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

State the boundary condition for cantilever beamDetermine the slope and deflection in every loaded situation as below:Cantilever beam, with concentrated load at the end.Cantilever beam, with a uniformly distributed load.Cantilever beam, with concentrated load at any point.

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Case 3: Cantilever beam, with a point load at end (Boundary conditions included)

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

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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 into eq. (a)

EI d2y = -Fx dx2

Integrate once we get;

EI dy = -Fx2 + A dx 2

Integrate once again we get;

EIy = -Fx3 + Ax + B 6

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a) To find the variables A and B

EI dy = -Fx2 + A .. (Slope Eq.) dx 2

When; x = L , dy/dx = 0

Thus;EI (0) = -F(L)2 + A 2

A = FL2 2

Boundary Conditions*

EIy = -Fx3 + Ax + B (Deflection Eq.) 6

Substitute A = FL2 2

EIy = -Fx3 + FL2x + B 6 2

When; x = L, y = 0

Thus;EI (0) = -F(L)3 + FL2(L) + B 6 2 B = -FL3 3EI*

Therefore, the COMPLETE slope and deflection equations are;

i) EI dy = -Fx2 + FL2 . (Slope Eq.) dx 2 2

ii)EIy = -Fx3 + FL2x - FL3 (Deflection Eq.) 6 2 3

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b) To find the Maximum slope and deflection

i)Maximum Slope

EI dy = -Fx2 + FL2 dx 2 2

When; x = 0,

EI dy = -F(0)2 + FL2 dx 2 2

dy = - FL2dx 2EI

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ii) Maximum deflection

EIy = -Fx3 + FL2x - FL3 6 2 3

When;x = 0

EIy = -F(0)3 + FL2 (0) FL3 6 2 3

EIy = - FL3 3

y = - FL3 3EI

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On the x-x section (distance x is based on point A), it is known that the moment at this section is;

M xx = -Fx2 2

Case 4: Cantilever beam, with a point load at end (Boundary conditions included)xA*

Solution

Substitute M = -Fx2 /2

EI d2y = -Fx2 dx2 2

Integrate once we get;

EI dy = -Fx3 + A dx 6

Integrate once again we get;

EIy = -Fx4 + Ax + B 24

a) Find the variable A and B

When ,x = L, dy/dx = 0

EI dy = -Fx3 + A dx 6 EI(0) = -F(L3) + A 6

A = FL3 6

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When , x = L, y = 0

EIy = -Fx4 + FL3x + B 24 6

EI(0) = -F(L)4 + FL3(L) + B 24 6

B = -FL4 8Therefore the complete slope and deflection equation are;

i) Slope EI dy = -Fx3 + FL3 dx 6 6

ii) DeflectionEIy = -Fx4 + FL3x FL4 24 6 8

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b) To find the Maximum slope and deflection

i)Maximum Slope

EI dy = -Fx3 + FL3 dx 6 6

When; x = 0,

EI dy = -F(0)3 + FL3 dx 6 6

dy = FL3dx 6EI

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ii) Maximum deflection

EIy = -Fx4 + FL3x FL4 24 6 8

When;x = 0

EIy = -F(0)4 + FL3(0) FL4 24 6 8

y = - FL4 8EI

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Case 5: Cantilever beam, with a point load NOT at the end

i) Max slope; dy = Fa2 dx 2EI

ii) Max deflection;y = Fa2 (3L-a) 6EIL*

Exercise1.A cantilever beam is 6m long and has a point load of 20 KN at the free end. The flexural stiffness is 110MNm2. Calculate the slope and deflection at the free end. (Answer: 3.27 x 10-3, 13mm)

2.A cantilever beam is 5m long and has a point load of 50KN at the free end. The deflection at the free end is 3mm downwards. The modulus of elasticity is 205GPa. The beam has a solid rectangular section with a depth 3 times the width.(D=3B). Determine;

i) the flexural stiffness, EI. (Answer: 694.4MNm2)ii)the dimension of the section. (Answer:D=591mm, B=197mm)

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3. A cantilever beam is 6m long with a UDL of 1 KN/m. the flexural stiffness is 100MNm2. Calculate the slope and deflection at the free end. (Answer: 360 x 10-6, 1.62mm)

4. A cantilever beam is 5m long and carries a UDL of 8KN/m. The modulus of elasticity is 205GPa and beam is a solid circular section. Calcuate;

i)The flexural stiffness which limits the deflection to 3mm at the free end. (Answer: 208.3MNm2)ii)The diameter of the beam. (Answer: 379mm)*

5. Based on the diagram below, calculate the maximum slope and deflection. The flexural stiffness is 300MNm2. (Answer: )

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