Question

Use bisection method to determine the point of maximum deflection of the beam subject to a linearly increasing distributed load shown in the figure below (the value of x where dy/dx= 0). Then substitute this value into the equation to determine the value of the maximum deflection. Use the following parameter values in your computation: L = 600 cm, E=50,000 kN/cm², I=30,000 cm⁴, and w₀ =1.75 kN/cm. 

Answer 1

Solution:- the values and figure(diagram) given in the question are as follows:

length of beam(L)=600 cm

E=50000 kN/cm^2

I=30000 cm^4

Wo=1.75 kN/cm

deflection equation-

$y=(W_{o}/120EIL)(-x^5+2L^2x^3-L^4x)$ , [Eq-1]

where, y=deflection

Wo

for maximum deflection, dy/dx=0

$dy/dx=(W_{o}/120EIL)(-5x^4+2L^2*3x^2-L^4)$

$dy/dx=(W_{o}/120EIL)(-5x^4+6L^2x^2-L^4)$

$0=(W_{o}/120EIL)(-5x^4+6L^2x^2-L^4)$

put value of L=6 m in above equation

$0=(W_{o}/120EIL)(-5x^4+216x^2-1296)$

$0=(-5x^4+216x^2-1296)$

$f(x)=-5x^4+216x^2-1296=0$ , [Eq-1]

Calculating root of above function by bisection method-

f(3)=-405+1944-1296=+243

f(2)=-80+864-1296=-512

root of above function in between 3 and 2

first approximation-

x1=(3+2)/2=2.5

f(2.5)=-195.3125+1350-1296=-141.3125

root of above function in between 3 and 2.5

second approximation-

x2=(3+2.5)/2=2.75 m

f(2.75)=+51.5429

root of above function in between 3 and 2.75

third approximation-

x3=(2.5+2.75)/2=2.625 m

f(2.875)=-45.028

root of above function in between 3 and 2.625

fourth approximation-

x4=(2.625+2.75)/2=2.6875 m

f(2.687)=+2.8735

root of above function in between 2.6875 and 2.625

fifth approximation-

x5=(2.625+2.6875)/2=2.65625 m

f(2.65625)=-20.88

root of above function in between 2.6875 and 2.65625

sixth approximation-

x6=(2.65625+2.6875)/2=2.6718 m

f(2.6718=-8.872

root of above function in between 2.6875 and 2.6718

seventh approximation-

x7=(2.6718+2.6875)/2=2.67965=2.68 m

f(2.68)=-2.5359

root of above function in between 2.6875 and 2.68

eighth approximation-

x8=(2.68+2.6875)/2=2.6837 m

f(2.6837)=+0.3233

root of above function in between 2.6837 and 2.68

ninth apprximation-

x9=(2.68+2.68375)/2=2.6835

f(2.6835)=+0.168

value of 0.16 ia very near to zero so the root of function is 2.6835

the maximum deflection occur at a distance 2.6835 m from left side

calculating maximum deflection(y_max)-

put values in above equation-(1)

for calculating maximum deflection put x=268.35 cm in above equation-(1)

y_max={1.75/(120*50000*30000*600)}*{-268.35^5+2*600^2*268.35^3-600^4*268.35}

y_max=-0.360633 cm

[negative sign represent deflection is vertically downward]

maximum deflection(y_max)=-0.360633 cm

[Ans]