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Carbon Steel Mechanical Design

This document contains calculations related to the mechanical properties of carbon steel, blade design parameters, and structural analysis of a wind turbine blade. It lists the elastic modulus and Poisson's ratio of carbon steel. It then details calculations for determining the maximum torque and bending moment experienced by different sections of a wind turbine blade based on given specifications. Finally, it performs calculations to determine the minimum required diameter of the blade based on strength and stiffness considerations.

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sam afi
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81 views4 pages

Carbon Steel Mechanical Design

This document contains calculations related to the mechanical properties of carbon steel, blade design parameters, and structural analysis of a wind turbine blade. It lists the elastic modulus and Poisson's ratio of carbon steel. It then details calculations for determining the maximum torque and bending moment experienced by different sections of a wind turbine blade based on given specifications. Finally, it performs calculations to determine the minimum required diameter of the blade based on strength and stiffness considerations.

Uploaded by

sam afi
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Calculations

At temperature of 25oC

Mechanical properties Carbon steel

Elastic modulus 190-210 GPa

Poison ratio 0.27-3.0

Tensile strength 276 – 1882 MPa

Yield strength 186 – 758 MPa

 Wblade = 8KN
 Rpm = 45
 Power = 239.9 kW/h
 Efficiency = 40%
 Length of blade = 35 m

Design based on strength:

η = Pelec / Pmec
0.4 = 239.9 / Pmec
Pmec = 600 kW/h
T = Pmec / ω
T = 600 x 103 / 45 x 2π / 60
T = 33.71 x 103 Nm 8kN

R1x = -1.23kN
R2x = 9.231kN
A B C
∑ MA = 0

1.3m 0.2m
R2y
R1y
1.23k x =M…………… 1

1.23k (1.3 + x) – 9.23k x =M…………….2

From 1 and 2

Mmax = 1.6kNm…………. at B

1.44
4443
3224
4444
444

R1x = 0.222 kN

R2x = -1.666 KN

∑ MA = 0

Mbx = 0.2886 kNm

MB = (0.28862 + 1.62)1/2

MB = 1.636 kNm

d =(16((1.6 x 103)2 + (33.71 x 103)2)1/2/2 x 700 x 106 / 6)1/3

d = 114mm

Design based on stiffness:

T = 33.71 x 103 Nm

Since F2/F1 = 5………………….3

Force responsible for torsion moment

Fn = F2 – F1…………………………….4

Fblade = F2 +F1…………………..5

From 3, 4 and 5

Fblade = 1.5 Fn

Fn = T/r
Fn = 33.71 x 103/ 35

Fnx = 0.963 kN

Fbladex = 1.444 kN

As , MB = 1.626 kNm

Sy = 700 MPa

Se’ = 0.5 Sut

Se ‘= 0.5 x 700

Se’ = 350 MPa

Se = CloadCsizeCsurfCtempCreliaSe’

Assumed surface is ground, reliability is of 50% and Load is bending.

Se = 1 x 1 x 0.767 x 1 x 1 x 350

Se = 268.45 MPa

For notch = 0.5mm ………………. Assumed

From fig. 6.36

q = 0.7 (for bending)

q = 0.74 (for torsion)

Kt = 3.5……………..assumed

Kf = 1+q(Kt-1)

Kf = 1+0.7(3.5-1)

Kf = 2.85

As, Kf│σmax│< Sy

So we use Kfsm = Kf = 2.85

Assuming the torque is steady

Using eq 9.6

dB = {32Nf [(Kf x M / Sf)2 +3(Kfsm x T/Sy)2]1/2/ π}1/3

dB = 150mm

Now, for unsteady torque but mean torque is equal to fluctuating torque
dB = {32Nf {[(Kf x Mm / Sf)2 +3(Kfsm x Tm/Sy)}2+[(Kf x Mm / Sf)2 +3(Kfsm x Tf/Sy)}2]1/2}/ π}1/3

dB = 170mm

Angular deflection;

Ф = (TL) / (JG)

G = E / 2(1+v)

G = 77.51GPa

Ф = 2.735 x 10-3 rad.

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