# Chapter6-Beams and bending part 3¶

## Ex1-pg215¶

In [1]:
import math
#calculate The deflection at the free
l = 5.;## feet
W = 150.;## lb
w = 120.;## lb. per foot run
l1 = 3.;## feet
b = 3.;## inches
d = 6.;## inches
E = 1.5*10**6;## lb/in^2
I = (1./12.)*b*d**3;## in^4
y_B1 = (W*l**3)/(3.*E*I);## feet
y_B2 = (w*l1*l1**3)/(8.*E*I) + (l-l1)*(w*l1*l1**2)/(6.*E*I);## feet
y_B = (12**3)*(y_B1+y_B2);## inches
print'%s %.4f %s'%('The deflection at the free end =',y_B,'inches')

The deflection at the free end = 0.1823 inches


# Ex2-pg218¶

In [3]:
import math
b = 4.;## inches
d = 9.;## inches
l = 12.;## feet
y_c = 1/4.;## inches
E = 1.5*10**6;## lb/in^2
I = (1/12.)*b*d**3;## in^4
W = y_c*384.*E*I/(5.*12**3.*l**3);## inches
print'%s %.d %s'%('Uniform distributed load, the beam should carry is, W =',W,'lb-wt');

##there is an error in the answer given in text book

Uniform distributed load, the beam should carry is, W = 2343 lb-wt


## Ex3-pg224¶

In [4]:
import math
#calculate 'Uniform distributed load the beam should carry
b = 4.;## inches
d = 9.;## inches
l = 12.;## feet
y_c = 1/4.;## inches
E = 1.5*10**6;## lb/in^2
I = (1/12.)*b*d**3;## in^4
W = y_c*384.*E*I/(5.*12**3.*l**3);## inches
print'%s %.d %s'%('Uniform distributed load, the beam should carry is, W =',W,'lb-wt');

##there is an error in the answer given in text book

Uniform distributed load, the beam should carry is, W = 2343 lb-wt


## Ex4-pg225¶

In [5]:
import math
#calculateThe deflection for a uniformly distributed load
d = 6.;## feet
l = 60.;## feet
f = 15./2.;## tons/in^2
E = 13000.;## tons/in^2
k1 = 2.*f/(12.*d);## k1 = M_r/I
k2 = k1/(l*12./8.);##k2 = W/I
y_c = (5./384.)*k2*l**3 *12**3 /E;## inches
##If the giredr is of constant deapth and uniform strength, it bends to an arc of a circle of radius R
R = E*d*12/(2*f);## inches
delta = (l*12)**2 /(8*R);## inches
print'%s %.2f %s'%('The deflection for a uniformly distributed load on it is,delta =',delta,'inches');

The deflection for a uniformly distributed load on it is,delta = 1.04 inches


## Ex5-pg227¶

In [7]:
import math
#calculate The ratio of deapth to span
f = 8.;##tons/in^2
E = 12800.;## tons/in^2
k1 = 1./480.;##central deflection = k = delta/l
k2 = (5./24.)*(f/E)/k1 ;##k2 = d/l =  deapth to span ratio
print'%s %.3f %s'%('The ratio of deapth to span, d/l =',k2,'');

The ratio of deapth to span, d/l = 0.062


## Ex6-pg228¶

In [11]:
import math
#calculate 'A section with d and If the deapth of section is limited
w = 550.;## lb. per foot run
f = 1000.;## lb/in^2
l = 20.;## feet
d_limit = 15.;## inches
E = 1.5*10**6;## lb/in^2
##central ddeflection
delta = (1./2.);## inches
d = (5./24.)*(f/E)*20.*12./(1./(2.*20.*12.));## inches
M = w*l*l*12./8.;## lb-inches
b = M/(f*(1./6.)*d**2);## inches
print'%s %.d %s'%('A section with d =',round(d),'inches')
print'%s %.d %s'%('b =',round(b),'inches will do.')
f1 = (1./(2.*20.*12.))*(d_limit/(l*12.))*E/(5./24.);## lb/in^2
b = M/(f1*(1./6.)*d_limit**2);## inches
print'%s %.d %s'%('If the deapth of section is limited to',d_limit,'inches,then')
print'%s %.1f %s'%('f =',f1,'lb/in^2')
print'%s %.1f %s'%('b =',b,'inches');

##tha answer is correct only, but it is approximated in the text book.

A section with d = 16 inches
b = 8 inches will do.
If the deapth of section is limited to 15 inches,then
f = 937.5 lb/in^2
b = 9.4 inches


## Ex7-pg235¶

In [8]:
import math
#calculate The position of maximum deflection occurs and  The maximum deflection and The deflection at the centre
l = 20.;## feet
b = 4.;## feet
W = 5.;## tons
d = 12.;## inches
h = 5.;## inches
I_xx = 220.;## in^4
E = 13000.;## tons/in^2
a = l-b;## feet
##for maximum deflection
x = math.sqrt((a**2 + 2.*a*b)/3.);## feet
y_max = x*12**3 *((a**2 + 2.*a*b) - x**2)/(6.*E*I_xx);## inches
##for deflection at the centre
x1 = 0.5*l;## inches
y_x1 = x1*12**3 *((a*82. + 2.*a*b) - x1*82.)/(6.*E*I_xx);## inches
print'%s %.2f %s'%('The position of maximum deflection occurs at x =',x,'feet')
print'%s %.4f %s'%('The maximum deflection is, y_max =',y_max,'inches');
print'%s %.3f %s'%('The deflection at the centre,',y_x1,'inches');

The position of maximum deflection occurs at x = 11.31 feet
The maximum deflection is, y_max = 0.2917 inches
The deflection at the centre, 0.624 inches


## Ex8-pg236¶

In [10]:
import math
#calculate The position of the maximum deflection and the maximum deflection
d = 12.;## inches
h = 5.;## inches
l = 20.;## feet
E = 13000.; ##tons/in**2
I_xx = 220.;## in**4
W = 4.;## tons
W1 = 3.;## tons
a = 15.;## feet
b = l-a;## feet
a1 = 16.;## feet
b1 = l-a1;## feet
K1 = (-2*W1*b1*l)/(W1*b1-W*b);
K2 = (W*b*a**2 + 2*a*W*b**2 + 2*W1*b1*l**2 - W1*b1*a1**2 -2*W1*a1*b1**2 +W1*b1*l**2)/(3*(W1*b1 - W*b));
x = -0.5*K1 + math.sqrt(-K2 + 0.25*K1**2);## feet
x1 = l-x;## feet
y_max = W*b*x*1728.*(a**2 +2*a*b -x**2)/(6.*E*I_xx*l) + W1*b1*x1*1728.*(a1**2 +2.*a1*b1 -x1**2)/(6.*E*I_xx*l);## inches
print'%s %.2f %s'%('The position of the maximum deflection is, x =',x,'feet.');
print'%s %.4f %s'%('And the maximum deflection is, y_max =',y_max,'inches');

The position of the maximum deflection is, x = 10.26 feet.
And the maximum deflection is, y_max = 0.4488 inches


## Ex9-pg243¶

In [11]:
import math
#calculate The position of the maximum deflection and the maximum deflection
b = 18.;## inches
d = 7.;## inches
w1 = 1.;## ton per foot run
w2 = 3.;## ton per foot run
I_xx = 1149.;## in**4
E = 13000.;## tons/in**2
R_A = 0.5*b + (b/3.);## tons
R_B = 0.5*b + (2.*b/3.);## tons
##integrating M = E*I*y'', to get E*I*y' and making y' = 0;, we get maximu deflection
x = 9.18;## by trial and error method
y_derivative = -R_A*0.5*x**3 + x**4 /6. +0.5*(2./3.)*(1./b)*(1./4.)*x**5 + 469.8;
y = -R_A*0.5*x**3 /3. + x**4 /24. +0.5*(2./3.)*(1./b)*(1./(4.*5.))*x**5 + 469.8*x;
y_max = y;## inches
print'%s %.2f %s'%('The position of maximum deflection from the end A, x =',x,'inches')
print'%s %.4f %s'%('Maximum deflection, y_max =',y_max*12**3 /(E*I_xx),'inches')

The position of maximum deflection from the end A, x = 9.18 inches
Maximum deflection, y_max = 0.3164 inches


## Ex10-pg254¶

In [13]:
import math
#calculate The position of the maximum deflection and the maximum deflection and  stress induced
b = 18.;## inches
d = 6.;## inches
l = 16.;## feet
W = 2.;## tons
h = 1./2.;## inches
I_xx = 841.76;## in**4
E = 13000.;## tons/in**2
P = W + math.sqrt(2.*W*h*48.*E*I_xx/(l*12.)**3 + 2.*W);## tons
M_max = P*l*12./4.;## ton-inches
Z = 2.*I_xx/b ;## in**3
f = M_max/Z ;## tons/in**2
delta = P*(l*12.)**3 /(48.*E*I_xx);## inches
print'%s %.3f %s'%('The maximum instantaneous deflection delta =',delta,'inches')
print'%s %.3f %s'%('stress induced, f =',f,'tons/in**2');
##there is an error in the answer given in text book

The maximum instantaneous deflection delta = 0.193 inches
stress induced, f = 7.362 tons/in**2


## Ex11-pg262¶

In [17]:
import math
#calculate Number of plates required and Number of plates required and The initial radius to which the plates must be bent
l = 3.;## feet
b = 3.;## inches
t = 3./8.;## inches
W = 1500.;## lb.
f = 12.;## tons/in**2
E = 30.*10**6;## tons/in**2
M_max = W*l*12./4. ;## lb-inches
M_r = f*(1./6.)*b*t**2 *2240.;## lb-inches
n = M_max/M_r ;## no. of plates
n = round(n+1);
f = M_max/(n*(1./6.)*b*t**2);## lb/in**2
R = E/(2.*f/t) ;## inches
delta = (l*12.)**2 /(8.*R);## inches
print'%s %.d %s'%('Number of plates required, n =',n,'');
print'%s %.4f %s'%(' The central deflection, delta =',delta,'inch.');
print'%s %.3f %s'%('The initial radius to which the plates must be bent, R =',R,'inches');

Number of plates required, n = 8
The central deflection, delta = 0.6912 inch.
The initial radius to which the plates must be bent, R = 234.375 inches