Ax Ma 13656.25 N P=WxL P = 3656.25 N W = 0.5 x 1.2x (25)^2 x 1.00 W = 375 N/m LENGTH = 9.75M 1828.125 N SHEAR FORCE DIAGRAM N LENGTH = 4.875 M LENGTH = 4.875 M 02 BENDING FORCE DIAGRAM 4.456.055 N/m 02 JON/m Fx =0 Ax=0 ' t ) m x + m > Fy=0 Ay - 3656.25 N = 0 Ay = 3656.25 N Ma=0 Ma - 3656.25 N x 4.875 m = 0 Ma = 17,824.219 N/m Fy=0 3656.25 N - 375 x 4.875 - V V = 1828.125 N m = 0 m+Ma=0 m=-17,824.219 N/m A m = area a = 0.5x (3656.25 + 1828.125) x 4.875 a = 13,368.164 Nm m = -17,824.219 + 13,368.164 = 4,456.055 N/m = 1 Distributed load on blades = × Pair X wind² x width of the blade Pair = 1.2 kg/m² ma Table 1: Classification of wind speed according to Beaufort Scale[1] Wind speed Wind force m/s from to acc. to Beaufort Wind force Visible effects inland notation 0.0 0.2 0 Calm Smoke rises vertically 0.3 1.5 1 Light air 1.6 3.3 2 Light breeze 3-4 5.4 ま 3 Gentle breeze 5.5 7-9 4 Smoke indicates wind, wind vanes do not Wind perceptible on face, wind vanes move move Leaves and thin branches move, wind extends pennants Moderate breeze Thin branches move, dust and paper are raised 8.0 10.7 5 Fresh breeze Small trees begin to sway, white caps form on lakes 10.8 13.8 6 Strong breeze Thick branches move, telegraph lines whistle 13.9 17.1 7 Moderate gale 17.2 20.7 8 Fresh gale Whole trees move, difficult to walk Wind breaks branches off trees 20.8 24.4 9 Strong gale Minor damage to houses (roof tiles) 24.5 28.4 10 Whole gale Trees are uprooted 28.5 32.6 11 Storm 32.7 56 12-17 Hurricane storm damage Significant damage to houses (very rare inland) Widespread devastation

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Hello. For my assignment I have ot work out the bending and shear for a wind turbine (this is just focusing on the blade). I just wanted to check if my diagram was correct per these measurements. below are my calculations... "

Per the image above, we have defined the optimised width and depth to be:

Width b = 1.0 m and depth t = 0.05 m.

The formula for moment of inertia for a rectangular cross section is shown below:

Therefore, to calculate the bending stress is shown below:

Thereby, the calculated bending stress is below the lower yield strength limit of aluminium alloy (109 MPa) and confirms the blade is capable of withstanding the bending forces. However as 43.61 MPa is slightly higher than the safety factor of 3, it is within a safety factor of 2 which is acceptable.

The shear stress is calculated using the following equation:

Where:

V max is the maximum share force (3729.39 N)
Q is the first moment of area about the neutral axis.
t is the thickness if the blade

As the fatigue strength ranges between 68.2MPa and 169Mpa, the calculated shear stress is below the safety factor of the shear strength for this aluminium alloy." can you please review and check if i am correct or incorrect in my solution.