# Calculating the Second Moment of Area

It is necessary to design a square-section composite bar that will withstand an applied load without deflecting more than a specified amount.The design specification for the bar, including its dimensions and loading criteria, is given below.

Using the formula below, calculate the second moment of area of the bar and use this to estimate the modulus of a composite material that will just meet the design specification.

The composite is made up of polymer fibres all aligned along the length of the bar combined with an epoxy matrix and each has the properties listed below. Estimate the minimum volume fraction of fibres that the composite must contain to meet the design specification.

Length: l= 0.5m

Thickness: a=10mm

Max deflection=2mm

Applied load: F=5kg at centre of the bar

Acceleration due to gravity:g =9.8m s(to power -2)

Second moment of area of a square section of side a: I=a(to power of 4)/12

Max deflection of beam of material modulus E=Fl(to power of 3)/48EI

Modulus of fibres: E fibre= 125GN m(to power -2)

Modulus of matrix: E epoxy= 3.5GN m(to power -2)

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#### Solution Preview

Let's denote with (E) the necessary modulus for the composite beam so that it withstands the load (F) getting the maximum acceptable deflection (derivative_max) .

Of ...

#### Solution Summary

This complete solution includes formatted calculations and explanation in the attached Word document. 447 words.

Ideal solutions concerning rotating solid and hollow shafts

3a.

A solid shaft of diameter 30mm rotating at 1000rpm delivers a power of 25kW determine (assume Shear Modulus for the shaft, G=80 GPa)

The Shear Stress in the shaft

The angular twist of the shaft as a result of the Shear Stress from part (i)

The important parameters to note here are

Power, P=25kW=25,000W

Diameter of shaft (needs to be in metres), D=30mm=0.03m

Radius of shaft, r=D/2=0.015m

Rotational speed (needs to be in radians per second),

ω=2π×1000rpm=2000π rads/s

Shear Modulus, G=80 GPa=8×〖10〗^10 Pa

(i) First we need to use the Power-Torque equation {1} to determine the Torque in the shaft

P=ωT {1}

So re-arranged to derive the Torque in the shaft

T=P/ω=25,000/2000π=3.978 N.m

We now need to use the equation {2} for the polar second moment of area,

〖 J〗_solid for a solid shaft

J_solid=(πD^4)/32 {2}

J_solid=(π(0.03)^4)/32=7.95×〖10〗^(-8) m^4

One now knows T,J_solid and shaft radius r so we use the Torque-Shear

Stress, equation {3} to determine the maximum Shear stress, τ in the shaft

T/J_solid =τ/r {3}

τ=Tr/J_solid =(3.978×0.015)/(7.95×〖10〗^(-8) )=7.51×〖10〗^5 N.m^(-2)=751 kN.m^(-2)

(ii) For the angle of twist we need to use the Angle (in radians)-Shear stress equation{4} to determine angular twist in the shaft over length, L , for Shear Modulus, G=80 GPa

θ=τL/Gr {4}

θ=(7.51×〖10〗^5×2)/(8×〖10〗^10×0.015)=1.25×〖10〗^(-3) rads

You can convert to degrees if you want noting the conversion factor (2π= 360^0

θ=(360/2π)1.25×〖10〗^(-3)=0.072^0

3b.

Find the diameter of a solid shaft resulting in a transmission Torque of 50 kNm for a maximum allowable Shear Stress in the shaft of 100 MPa

The important parameters to note here are

Transmission Torque, T=50kNm=50,000 Nm

Maximum allowable Shear stress in shaft τ=100 MPa=〖10〗^8 Pa

We need to use the equation {2} for the polar second moment of area,

J_solid for a solid shaft

J_solid=(πD^4)/32 {2}

And equation {3} relating Shear stress to Torque and radius of shaft

T/J_solid =τ/r {3}

Noting that radius r=D/2 and substituting {2} in {3} we get

T/((πD^4)⁄32)=τ/(D⁄2) {5}

Simplifying {5}

32T/(πD^4 )=2τ/D

32T/(πD^3 )=2τ {6}

Rearranging {6} to make D the subject

D^3=16T/πτ {7}

Putting in given values to {7} we get

D^3=(16×50,000)/(π×〖10〗^8 )=2.55×10^(-3) m^3

D=∛(2.55×〖10〗^(-3) )=13.7 cm

3c.

Calculate the power delivered by a rotating hollow shaft with an outer diameter of 150mm, an inner diameter of 100mm, rotating at a speed of 3 rps for maximum Shear Stress of 50 MPa

The important parameters to note here is that the shaft is hollow so we need to use equation {8} defining the polar second order of area for a hollow tubular shaft of outer diameter D and inner diameter d

J_tube=π(D^4-d^4 )/32 {8}

We are told outer diameter, D=150mm=0.15m

Inner diameter, d=100mm=0.1m

Using {8} then

J_tube=π(〖0.15〗^4-〖0.10〗^4 )/32=4×〖10〗^(-5) m^4

We can now use equation {3} to work out the maximum Torque, T on the shaft for maximum Shear stress, τ=50MPa=5×〖10〗^7 Pa

T/J_tube =τ/r {3}

T=(τJ_tube)/r

Putting in values noting that the value radius value, r in the above relates to the radius of the outer diameter which is r=D/2=0.15/2=0.075m

Maximum Torque T=(5×〖10〗^7×4×〖10〗^(-5))/0.075=2.67×〖10〗^4 N.m

We now can use equation {1} to determine the maximum power delivered by this shaft where the given angular velocity of shaft is ω=2πN and where N=3 rps

P=ωT {1}

P=2π×3×2.67×〖10〗^4=503.3kW