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현재 위치

Simple beam, distributed load, and deflection at a distance from a point

작성자 Uploader : metalheart 작성일 Upload Date: 2019-11-23변경일 Update Date: 2019-12-16조회수 View : 336

As shown in the figure, find the deflection of the beam vx (mm) when the load w (kN / m) is applied to a simple beam of length L (m).

The reaction force at points A and B is RA = RB = w * L / 2.

If we find the moment Mx (kN-m) at the point x (m) distant from the point A,

Mx = RA*x-w*x*(x/2) = (w/2)(L*x-x^2)

Since EI*v″= Mx, if you integrate it twice,

EI*v = (w/2)∬L*x-x^2 dx

= (w/2)∫(L/2)*x^2 - (1/3)*x^3 + C1 dx

= (w/2)*((L/6)*x^3 - (1/12)*x^4 + C1*x + C2)

At x = 0, vx = 0, thus, C2 = 0

At x = L, vx = 0, thus C1 = -(1/12)*L^3

EI*v = (w/2)*((L/6)*x^3 - (1/12)*x^4 - ((1/12)*L^3)*x )

vx = -(w*x/(24*EI))*(L-x)(L^2+L*x-x^2)

If you apply MPa for the unit of E and cm^4 for the unit of I,

((kN/m)*m/(1000*kN/m^2*(0.01m)^4))*m^3 = 10^5*m

If the unit of vx is expressed in mm,

vx = -(w*x/(24*EI))*(L-x)*(L^2+L*x-x^2)*10^8

*** 참고문헌[References] ***

vx = -(w*x/(24*E*I))*(L-x)*(L^2+L*x-x^2)*10^8

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현재 위치

Simple beam, distributed load, and deflection at a distance from a point

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