DEFLECTION OF BEAMS

DEFLECTION OF BEAMS

The following are the standard cases for deflection of beams-

CaseSlopeDeflection

Cantilever with Point Load

W

L

W must be in Newtons S = W.L2 2E(Will be 0 at support and Max at load end = W.L3 3E(Will be 0 at support and Max at load end

Cantilever with UDL

UDL of ( N/m

L

UDL of ( N/m S = (.L3 6E(Will be 0 at support and Max at load end = (.L4 8E(Will be 0 at support and Max at load end

Simply Supported with Load at centre

W

L

RA RB

W must be in Newtons S = W.L2 16E(Will be Max at ends = W.L3 48.E(Will be Max at centre

Simply Supported with UDL

UDL of ( N/m

L

RA RB

UDL of ( N/m S = (.L3 24E(Will be Max at ends = 5. (.L4 384.E(Will be Max at centre

Built in with central Point Load

W

L

W must be in Newtons S = 0Will be 0 at ends and centre = __W.L3 192.E(Will be Max at centre

Built in with UDL

UDL of ( N/m

L

UDL of ( N/m S = 0 Will be 0 at ends and centre = _ (.L4 384.E(Will be Max at centre

Worked Examples

A cantilever is made from 43 grade structural steel and is 3m long and made of 120mm tube with a wall thickness of 20mm ( 80mm bore ). It carries a load of 500kg at the end.

Calculate-

1) The deflection

2) Stress

3) The stiffness

4) The natural frequency _________________

This is the case below-

Cantilever with Point Load

W

L

W must be in Newtons S = W.L2 2E(Will be Max at B = W.L3 3E(Will be Max at B

For grade 43 steel, which is the normal Black Bar or hot rolled section, E will be 205GN/m2 ( 205 x 109 ) with a yield stress of 275MN/m2.

W is 500kg = 500 x 9.81 = 4905N

The first step is to calculate the second moment of area-

I = ( (D4 d4) = 3.142 ( 0.124 - 0.084 ) = 3.142 ( 0.0002074 0.00004096) 64 64 64

I = 3.142 ( 0.0001665 )

64

I = 0.00000817m4 or 8.17 x 10-6

1) For deflection Deflection ( = - W.L3 = -_______ 4905 x 33__________

3E( 3 x 205 x 109 x 8.17 x 10-6Deflection ( = -_______ 4905 x 33_____ = ___ 132435_____

3 x 205 x 8.17 x 109-6 5024.55 x 10-3Deflection ( = -____44145_____ = 0.026m or 26mm

5024.55 x 1000

The deflection at the end will be 26mm.2) Stress

Stress ( = M.y Where M = bending moment

( y = Radius

I = Second moment of area

The bending moment for a cantilever is given by W x L

Stress ( = M.y = ( W x L ). R = 4905 x 3 x 0.06 ( ( 8.17 x 10-6

Stress ( = 108000000N/m2 or 108MN/m2The yield stress is 275MN/m2 and we should not exceed 50 to 60% of yield stress depending on type of load 60% for dead of fixed load and 50% for live load or one which is loaded and unloaded. 50% of 275 is 137.5 MN/m2 and 60% is 165 MN/m2. We only have 108 MN/m2 so it is perfectly safe.

The Factor of safety is given by FOS = Yeild stress = 275 = 2.54

Load stress 108

A minimum factor of safety of 1.67 to 2 is normally considered as acceptable ( 1.67 for dead loads and 2 for live loads which correspond to 50 and 60% of yield.

3) Stiffness

The Stiffness is given by the formula-

Stiffness = __Load__ = 4905N = 188634N/m or 188.6kN/m

Deflection 0.026m

Stiffness is the theoretical load to cause a deflection of 1m. The reason I say theoretical is that the material will fail long before it can deflect by that amount.

4) Natural frequency

The natural frequency of any beam is given by the formula-

Natural Frequency (n = 1 __9.81___ 2( ( deflection

Natural Frequency (n = ___1____ __9.81_ 2 x 3.142 ( 0.026m

Natural Frequency (n = __1__ x ( 377.3 = 19.42 6.284 6.284

Natural Frequency (n = 3.09 cycles/second or Hertz

Example 2

A beam is 4m long and carries a point load in the centre of 2 Tonne and a UDL of 0.5Tonne/m. The beam is grade 43 steel and I section and 100mm wide at top, 160mm deep and 10mm thick. Calculate-

a) Deflection

b) Natural frequency.In this case, the loads are 2 Tonne = 2 x 1000 x 9.81 = 19620N

0.5 T/m = 0.5 x 1000 x 9.81 = 4905N/mThe first step is the second moment of area

I = BD3 bd3) = 0.1 x 0.163 - 0.09 x 0.143 ) = 0.0004096 - 0.0002469 12 12 12

I = 0.0001627

12

I = 0.0000135m4 or 13.6 x 10-6 m4

We can now divide it into two sections-

1) The point load

2) The UDLWork out the deflections and add them together to give the total deflection-

For Point Load

Deflection ( = - W.L3 = -_______ 19620 x 43__________

48E( 48 x 205 x 109 x 13.5 x 10-6Deflection ( = -_______ 19620 x 43_____ = ___ 1255680_____

48 x 205 x 13.5 x 109-6 132840 x 10-3Deflection ( = -____44145_____ = 0.00945m or 9.4mm

5024.55 x 1000

For UDL Deflection ( = -5 (.L4 = -___5 x 4905 x 44__________

384E( 384 x 205 x 109 x 13.5 x 10-6Deflection ( = -___ 24525 x 256_____

384 x 205 x 13.5 x 10-3 Deflection ( = 0.0059m or 5.9mmThe total deflection will be the two added together

(Beam = (Point Load + (UDL = 9.4mm + 5.9mm

(Beam = 15.3mm Natural frequency

The natural frequency of any beam is given by the formula-

Natural Frequency (n = 1 __9.81___ 2( ( deflection

Natural Frequency (n = ___1____ __9.81_ 2 x 3.142 ( 0.0153m

Natural Frequency (n = __1__ x ( 641.2 = 25.32 6.285 6.284

Natural Frequency (n = 4.03 cycles/second or Hertz

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