Question

5.13 Figure P5.13a shows a uniform beam subject to a lin- early increasing distributed load. The equation for the result- ing elastic curve is (see Fig. P5.13b) htm U) (P5.13) 3 3" Use bisection to determine the point of maximum deflection (i.e, the value of x where dy/dx 0). Then substitute this value into Eq. (P5 13) to determine the value of the maximum deflection. Use the following parameter values in your FIGURE P5.9 PROBLEMs 149 5.16 The resistivity p of doped silicon is based on the charge q on an electron, the electron density n, and the elec- tron mobility . The electron density is given in terms of the doping density N and the intrinsic carrier density n The electron mobility is described by the temperature T, the ref- erence temperature Tor and the reference mobility Ho. The equations required to compute the resistivity are where 2.42 Determine N, given To = 300 K, T = 1000 K, μ0= 1360 cm2 (Vs)", q = 1.7 x 10-19 C, n 621 x 10, crn_. and a desired p 6.5 x 10 V s em/C. Employ initial guesses of N-0 and 2.5 x 1010. Use (a) bisection and (b) the false position method 5.17 A total charge Q is uniformly distributed around a ring shaped conductor with radius a. A charge g is located at a distance x from the center of the ring (Fig. P5. 17). The force exerted on the charge by the ring is given by FIGURE P5.13 F=4me) (x2+a2)3/2 computation: L 600 cm. E = 50,000 kN/cm, 1 = 30,000 cm. and u'= 2.5 kN/cm 5.14 You buy a $35,000 vehicle for nothing down at $8,500 per year for 7 years. Use the bisect function from Fig. 5.7 to determine the interest rate that you are paying. Employ where o8.9 x 10CN m) Find the distance x where the force is 1.25 N if q and O are 2 x 10 C for a ring with a ius of d.85 m

I wonder how to a problem 5.13.
This problem is related to applied numerical methods with Matlab(third edition).

I want Matlab code.

Answer 1

Please find attached the MATLAB code for the above problem below:

L = 600;
E = 50000;
I = 30000;
w0 = 2.5;
tolerance = 1e-6;
% y function
y = @(x) (w0/(120*E*I*L))*(-x^5+2*(L^2)*(x^3)-(L^4)*x);
% derivative function
f = @(x) (w0/(120*E*I*L))*(-5*x^4+2*3*(L^2)*(x^2)-(L^4));

% start and end position of the beam
x1 = 0;
x2 = 600;

%bisection method
while (x2-x1)/x2>tolerance
   f1 = f(x1);
   f2 = f(x2);
   if f1*f2 > 0
       break;
   end
   x0 = (x1+x2)/2;
   f0 = f(x0);
   if f1*f0 < 0
       x2 = x0;
   else
       x1 = x0;
       f1 = f0;
   end
end
%result
fprintf('The maximum deflection of the beam occurs at x = %f and the deflection is %f\n',x0,y(x0));

%RESULT : The maximum deflection of the beam occurs at x = 268.328047 and the deflection is -0.515190