# Chapter5-Beams and bending¶

## Ex1-pg140¶

In [1]:
import math
#calculate maximum intensity of stress
b = 6.;## inches
t = 1./2.;## inch
R = 40.;## feet
E = 13000.;## tons/in^2
y = t/2.;## inch
f = (E/(R*12.))*(y);## tons/in^2
print'%s %.2f %s'%('The maximum intensity of stress induced is f =',f,'tons/in^2')

The maximum intensity of stress induced is f = 6.77 tons/in^2


## Ex2-pg141¶

In [2]:
import math
#calculate moment of resistance  and The radius to which it should be bent
d = 14.;## inches
I = 442.57;##inch units
f = 8.;## tons/in^2
E = 13000.;## tons/in^2
R = E*d/(2.*f);## inches
M_r = f*(I/(d/2.));## ton-inches
print'%s %.d %s'%('The radius to which it should be bent is R =',R,'inches')
print'%s %.1f %s'%('The radius to which it should be bent is R =',R/12,'feet');
print'%s %.1f %s'%('The moment of resistance is M_r =',M_r,'ton-inches');

The radius to which it should be bent is R = 11375 inches
The radius to which it should be bent is R = 947.9 feet
The moment of resistance is M_r = 505.8 ton-inches


## Ex3-pg141¶

In [1]:
import math
#calculate the load is uniformly spread over its span and  the load is concentrated at the centre
d = 16.;## inches
I = 618.;## inch units
l = 24.;## feet
f = 15./2.;## tons/in^2
Z = I/(d/2.);## inch-units
M_r = f*Z;## ton-inches
W1 = 8.*M_r/(12.*l);##tons
##If the load is concentrated at the centre,BM  = W*l/4
W2 = 4.*M_r/(12.*l);##tons
print'%s %.1f %s'%('If the load is uniformly spread over its span, then W is given by W1=',W1,'tons')
print'%s %.3f %s'%('If the load is uniformly spread over its span, then W is given by W1 =',W1/l,'ton per foot run');
print'%s %.2f %s'%('If the load is concentrated at the centre, then W is given by W =',W2,'tons')

If the load is uniformly spread over its span, then W is given by W1= 16.1 tons
If the load is uniformly spread over its span, then W is given by W1 = 0.671 ton per foot run
If the load is concentrated at the centre, then W is given by W = 8.05 tons


## Ex4-pg142¶

In [2]:
import math
#calculalte The maximum permissible span for this beam
d = 20.;## inches
I = 1673.;## inch units
W = 3./4.;## ton per foot run
f = 8.;## tons/in^2
Z = I/(d/2.);## inch-units
M_r = f*Z;## ton-inches
l = math.sqrt(M_r*32/(3*12));##feet
print'%s %.2f %s'%('The maximum permissible span for this beam is l =',l,'feet')

The maximum permissible span for this beam is l = 34.49 feet


## Ex5-pg145¶

In [ ]:



## Ex6-pg154¶

In [3]:
import math
#calculate X
w = 160.;## lb. per sq. foot
b = 3.;## inches
d = 9.;## inches
l = 15.;## feet
f = 1200.;## lb. per sq. inch
Z = (1./6.)*b*d**2;## in^3
M_r = f*Z;## lb-inches
x = M_r/(w*l**2*12./8.);## feet
print'%s %.1f %s'%('x =',x,'feet');

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

x = 0.9 feet


## Ex7-pg154¶

In [4]:
import math
#calculate area of cross section of beam
l = 20.;## feet
b = 9.;## inches
h = 10.;## feet
w = 120.;## lb. per cub. foot
f = 1100.;## lb/in^2
W = w*(3./4.)*l*h;## lb-wt
BM_max = W*l*12./8.;## lb-inches
##assumnig d = 2b
b = (6.*BM_max/(f*4.))**(1/3.);## inches
d = 2.*b;## inches
print'%s %.3f %s'%('b =',b,'inches')
print'%s %.2f %s'%('d =',d,'inches',)
print'%s %.d %s'%('A section',b,'')
print'%s %.d %s'%('X',d,'will therfore do.')

b = 9.030 inches
d = 18.06 inches
A section 9
X 18 will therfore do.


## Ex8-pg155¶

In [5]:
import math
#calculate beam of area
B = 5.;## inches
D = 12.;## inches
t1 = 0.55;## inches
t2 = 0.35;## inches
f = 15./2.;## tons/in^2
l = 16.;## feet
b = B-t2;## inches
d = D-2.*t1;## inches
I_xx = (B*D**3 - b*d**3)/12.;## in^4
Z = I_xx/6.;## in^3
M_r = f*Z;## ton-inches
W = M_r/(l*12./8.);## tons
w = W/l;## ton per foot run
print'%s %.2f %s'%('W =',W,'tons')
print'%s %.2f %s'%('w =',w,'ton per foot run')

W = 11.36 tons
w = 0.71 ton per foot run


## Ex9-pg156¶

In [6]:
import math
#calculate maximum stress
D = 19.5;## inches
d = 18.;## inches
l = 30.;## feet
t1 = 3./4.;## inch
rho1 = 450.;## lb. per cub. foot
rho2 = 62.5;## lb. per cub. foot
A = 0.25*math.pi*(D**2 - d**2);## sq. in
DW = rho1*l*A/144.;## lb-wt
WW = rho2*0.25*math.pi*(D-d)**2*l;## lb-wt
W = DW+WW;## lb-wt
BM_max = W*l*0.0004467202*12./8.;## ton-inches
I_xx = (math.pi/64)*(D**4 - d**4);## in^4
Z_xx = I_xx/(0.5*d+t1);## ton/in^2
f = BM_max/Z_xx;## ton/in^2
print'%s %.3f %s'%('The maximum stress f =',f,'ton/in^2');

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

The maximum stress f = 0.751 ton/in^2


## Ex11-pg158¶

In [7]:
import math
#calculate The maximum permissible span for this beam
b = 6.;## inches
d = 12.;## inches
t1 = 7./8;## inch
t2 = 1./2.;## inch
I_xx = (1./12.)*(b*d**3 - (b-t2)*(d-2*t1)**3);## in^4
Z1 = I_xx/b;## in ^3
A = 2*b*t1 + 0.5*(d-2*t1);## in^2
b = math.sqrt(A/2);## inches
d = 2.*b;## inches
Z2 = (1./6.)*b*d**2;## in^3
k = Z1/Z2 ;
print'%s %.2f %s'%('The ratio of strengths Z1/Z2 =',k,'');

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

The ratio of strengths Z1/Z2 = 4.24


## Ex12-pg159¶

In [8]:
import math
#calculate  the strength of the solid circular section is taken as unity
A = 15.625;## in**2
Z1 = 61.75;## in**3
Z2 = 14.63;## in**3
d = math.sqrt(4.*A/math.pi);## inches
Z3 = (math.pi/32.)*d**3.;## in**3
R1 = Z1/Z3;
R2 = Z2/Z3;
print('If the strength of the solid circular section is taken as unity')
print'%s %.2f %s %.2f %s'%('that of the rectangular section is',R1,' and of the I-section it is ',R2,'')

If the strength of the solid circular section is taken as unity
that of the rectangular section is 7.09  and of the I-section it is  1.68


## Ex13-pg161¶

In [9]:
import math
#calculate Position of the c.g of the sections
D = 8.;##h inches
B = 3.;## inches
t1 = 1/2.;## inch
t2 = 3/8.;## inch
b = B-t2;## inches
d = D-2.*t1;## inches
a1 = t1*B;## in**2
x1 = 0.5*B;## inches
a2 = t2*(D-2.*t1);## in**2
x2 = 0.5*t2;## inches
a3 = B*t1;## in**2
x3 = 0.5*B;## inches
a = a1+a2+a3;## in**2
P = (a1*x1+a2*x2+a3*x3)/(a1+a2+a3);## inches
I_xx = (1/12.)*(B*D**3. - b*d**3.);## in**4
I_AB = (1/3.)*t1*B**3. + (1/3.)*d*t2**3. + (1/3.)*t1*B**3.;## in**4
I_yy = I_AB - a*P**2.;## in**4
print'%s %.3f %s'%('Position of the c.g of the section P = ',P,' inches');
print'%s %.4f %s'%('I_xx = ',I_xx,'in**4')
print'%s %.3f %s' %('I_yy =',I_yy,'in**4');
##there is an error in the answer given in text book

Position of the c.g of the section P =  0.887  inches
I_xx =  52.9688 in**4
I_yy = 4.692 in**4


## Ex14-pg163¶

In [10]:
import math
#calculate The position of the c.g  and Maximum stresses induced
b = 6.;## inches
d = 4.;## inches
t = 5/8.;## inch
a1 = d*t;## in^2
y1 = d/2.;## inches
a2 = (b-t)*t;## in^2
y2 = t/2.;## inch
a = a1+a2;## in^2
J = (a1*y1+a2*y2)/(a1+a2);## inches
I_AB = (1/3.)*t*d**3 + (1/3.)*(b-t)*t**3;## in^4
I_xx = I_AB - a*J**2.;## in^4
I_yy = (1/12.)*t*b**3. + (1/12.)*(d-t)*t**3.;## in^4
print'%s %.1f %s'%('The position of the c.g is J = ',J,' inches');
print'%s %.2f %s %.2f %s'%(' I_xx =',I_xx,' in^4,'' I_yy = ',I_yy,'in^4');

##14(a)
H = 18.;## feet
l = 10.;## feet
w = 3/2.;## cwt/ per .sq. foot
y_c = 2.97;## inches
y_t = 1.03;## inches
W = (3/40.)*(w*l);## tons
BM_max = W*l*12./8.;## ton-inches
M_r = BM_max;## ton-inches
f_c = M_r*y_c/I_xx ;## tons/in^2
f_t = M_r*y_t/I_xx ;## tons/in^2
print'%s %.2f %s %.2f %s'%(' Maximum stresses induced are f_c = ',f_c,' tons/in^2,''f_t = ',f_t,' tons/in^2');

The position of the c.g is J =  1.0  inches
I_xx = 7.52  in^4, I_yy =  11.32 in^4
Maximum stresses induced are f_c =  6.66  tons/in^2,f_t =  2.31  tons/in^2


## Ex15-pg165¶

In [11]:
import math
#calculate The position of c.g is x_bar and Total uniformly distribute load over the span
b = 5.;## inches
d = 4.;## inches
t = 1/2.;## inches
a1 = b*t;## in**2
x1 = t/2.;## inches
y1 = b/2.;## inches
a2 = (d-t)*t;## in**2
y2 = t/2.;## inch
x2 = t + 0.5*(d-t);## inches
x_bar = (a1*x1+a2*x2)/(a1+a2);## inches
y_bar = (a1*y1+a2*y2)/(a1+a2);## inches
I_AB = (1/3.)*t*b**3. + (1/3.)*(d-t)*t**3.;## in**4
I_xx = I_AB - (a1+a2)*y_bar**2.;## in**4
I_yy = (1/3.)*t*d**3. + (1/3.)*(b-t)*t**3. - (a1+a2)*x_bar**2.;## in**4
print'%s %.3f %s %.2f %s'%('The position of c.g is x_bar = ',x_bar,'inches'and'y_bar = ',y_bar,'inches');
print'%s %.3f %s %.2f %s'%('I_xx = ',I_xx,'in**4'and 'I_yy = ',I_yy,' in**4');

##Example 15(a)
l = 12.;## feet
y_c = y_bar;##inches
y_t = b - y_c;##inches
f_t_max = 7.;## tons/in**2
f_c = y_c*f_t_max/y_t;## tons/in**2
M_r = f_t_max*I_xx/y_t;## ton-inches
W = M_r/(l*12./8.);## tons
print'%s %.2f %s'%(' Total uniformly distribute load over the span is W = ',W,' tons');

The position of c.g is x_bar =  1.074 y_bar =  1.57 inches
I_xx =  10.456 I_yy =  5.96  in**4
Total uniformly distribute load over the span is W =  1.19  tons


## Ex16-pg166¶

In [12]:
import math
#calculate The c.g of the section is y_bar and Load required
b1 = 6.;##inches
d1 = 1.;##inch
b2 = 9.;##inches
d2 = 1.;##inch
b3 = 10.;##inches
d3 = 2.;##inch
a1 = b3*d3;## in**2
y1 = d3/2.;## inches
a2 = b2*d2;## in**2
y2 = d3 + b2/2.;## inches
a3 = b1*d1;## in**2
y3 = b2+d3+d1/2.;## inches
y_bar = (a1*y1+a2*y2+a3*y3)/(a1+a2+a3);##inches
I_AB = (1/3.)*b3*d3**3 + (1/12.)*d2*b2**3. +b2*(d3+b2/2.)**2 + (1/12.)*b1*d1**3 + b1*(b2+d3+d1/2.)**2.;## in**4
I_xx = I_AB - (a1+a2+a3)*y_bar**2;## in**4
I_yy = (1/12.)*(d3*b3**3 + b2*d2**3 +d1*b1**3.);## in**4
print'%s %.3f %s'%('The c.g of the section is y_bar = ',y_bar,' inches');
print'%s %.2f %s %.2f %s'%(' I_xx = ',I_xx,' in**4' 'I_yy = ',I_yy,' in**4');

##Example 16(a)
l = 20.;## feet
y_t = y_bar;## inches
y_c = d1+b2+d3-y_t;## inches
f_t = 1.5;## tons/in**2
f_c = y_c*f_t/y_t;## tons/in**2
M_r = f_c*I_xx/y_c;## ton-inches
W = M_r*8./(l*12.);## tons
w = W/l;## ton per foot run
print'%s %.2f %s'%('Load required is w = ',w,' ton per foot run');

The c.g of the section is y_bar =  4.214  inches
I_xx =  640.06  in**4I_yy =  185.42  in**4
Load required is w =  0.38  ton per foot run


## Ex17-pg168¶

In [13]:
import math
#calculate stresses
b = 12.;## inches
d = 6.;## inches
h = 14.;## inches
t = 1/2.;## inch
A = 12.94;## in^2
##section moment of inertia
I_xx_s = 315.3;## in^4
I_yy_s = 22.27;## in^4

I_xx = 2.*I_xx_s + 2.*((1./12.)*h*(2.*t)**3 + h*2.*t*(d+t)**2.);## in^4
I_yy = 2.*(I_yy_s + A*(d/2.)**2.) + 2.*((1/12.)*2*t*h**3.);## in^4
print'%s %.2f %s %.2f %s'%('I_xx = ',I_xx,' in^4 I_yy = ',I_yy,' in^4');

I_xx =  1815.93  in^4 I_yy =  734.79  in^4


## Ex18-pg169¶

In [14]:
import math
b = 15.;## inches
d = 15/2.;## inches
h = 16.;## inches
t = 1/2.;## inch
P = 0.935;## inches
A = 12.33;## in**2
##section moment of inertia
I_xx_s = 377.;## in**4
I_yy_s = 14.55;## in**4

I_xx = 2.*I_xx_s + 2.*((1/12.)*h*(2.*t)**3. + h*2.*t*(d+t)**2.);## in**4
I_yy = 2.*(I_yy_s + A*(d/2. + P)**2.) + 2.*((1/12.)*2.*t*h**3.);## in**4
print'%s %.2f %s %.2f %s'%('I_xx = ',I_xx,'in**4 I_yy = ',I_yy,'in**4')

I_xx =  2804.67 in**4 I_yy =  1253.03 in**4


## Ex19-pg169¶

In [15]:
import math
#calculate maximum stresses
b1 = 16.;## inches
d1 = 6.;## inches
b2 = 9.;## inches
d2 = 7.;## inches
A = 14.71;## in**2
I_xx1 = 618.09;## in**4
I_yy1 = 22.47;## in**4
I_xx2 = 208.13;## in**4
I_yy2 = 40.17;## in**4
I_xx = I_xx1 + 2.*I_yy2;## in**4
I_yy = I_yy1 + 2.*(I_xx2 + A*(b2/2. + 0.5*0.4)**2.);## in**4
k_xx = math.sqrt(I_xx/(3.*A));## inches
k_yy = math.sqrt(I_yy/(3.*A));## inches
print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%('I_xx =',I_xx,' in**4 I_yy = ',I_yy,' in**4' 'k_xx =',k_xx,' inches'' k_yy = ',k_yy,' inches');

I_xx = 698.43  in**4 I_yy =  1088.62  in**4k_xx = 3.98  inches k_yy =  4.97  inches


## Ex20-pg170¶

In [16]:
import math
b1 = 7/2.;## inches
d1 = 7/2.;## inches
t1 = 3/8.;## inches
l = 18.;## inches
I_xx1 = 2.80;## in^4
I_yy1 = 2.80;## in^4
J = 1.;## inch
A = 2.49;## in^2
I_xx = 4.*(I_xx1 + A*(l/2. - J)**2);## in^4
k_xx = math.sqrt(I_xx/(4.*A));## inches
print'%s %.2f %s %.1f %s'%('I_xx = ',I_xx,' in^4'' k_xx = ',k_xx,' inches');

I_xx =  648.64  in^4 k_xx =  8.1  inches


## Ex21-pg170¶

In [17]:
import math
#calculate maximum stresses
b1 = 12.;## inches
d1 = 4.;## inches
A = 9.21;## in^2
I_xx1 = 200.1;## in^4
P = 1.055;## inches
I_yy1 = 12.12;## in^4
I_xx = 2.*I_xx1;
##for equal strength I_xx = I_yy
x = 2.*(math.sqrt(((I_xx/2.)-I_yy1)/A) - P);## in^4
print'%s %.2f %s'%('x = ',x,' inches');

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

x =  6.93  inches


## Ex22-pg184¶

In [18]:
import math
#calculate The moment of resistance offered by the flanges and The moment of resistance offered by the flanges
import scipy
from scipy import integrate
d =10.;## inches
b = 8.;## inches
t1 = 1.;## inch
t2 = 0.6;## inch
I = (1./12.)*(b*d**3. - (b-t2)*b**3.);## in^4
##(i) Resistance to  M
def fun(y):
x=(t2/I)*y**2.
return x

R1=R1[0]
##(ii) Resistance to F
def fun(y):
z=(4./I)*(25.-y**2.)
return z

R2=R2[0]
print'%s %.2f %s  %.2f %s  %.1f %s '%('The moment of resistance offered by the flanges is',1-R1,'M'           'The flanges take up ',(1-R1)*100,'percentage of the B.M.,''the web resisting only ',R1*100,'percentage of the B.M');
print'%s %.3f %s %.3f %s %.3f %s '%(' The shear borne by the web is',(1-2*R2),'F' and ' The web thus takes up ',(1-2*R2)*100,' ' and 'percentage of the shear force.'and  'the flanges resisting only',2*R2*100,' percentage of the shear force');

The moment of resistance offered by the flanges is 0.93 MThe flanges take up   92.71 percentage of the B.M.,the web resisting only   7.3 percentage of the B.M
The shear borne by the web is 0.894  The web thus takes up  89.362 the flanges resisting only 10.638  percentage of the shear force


## Ex23-pg186¶

In [19]:
import math
#calculate The maximum intensity of shear stress
b = 14.;## inches
d = 6.;## inches
t1 = 0.7;## inch
t2 = 0.4;## inch
F = 20.;## tons
I = (1./12.)*(d*b**3. - (d-t2)*(b-2.*t1)**3.);## in^4
q = F*t1*0.5*(0.5*b + (0.5*b-t1))/I;## ton/in^2
q_max = (F/(I*t2))*(d*t1*0.5*(0.5*b + (0.5*b-t1)) + t2*(0.5*b-t1)*(0.5*b-t1)*0.5);## tons/in^2
print'%s %.2f %s'%('The maximum intensity of shear stress is q_max = ',q_max,' tons/in^2');

The maximum intensity of shear stress is q_max =  4.09  tons/in^2


## Ex24-pg186¶

In [20]:
import math
#calculate The maximum intensity of shear stress
b = 4.;## inches
d = 13/2.;## inches
t = 1/2.;## inch
a = 4.;## inches
F = 10.;## tons
a1 = b*t;## in^2
y1 = t/2.;## inch
a2 = (d-t)*t;## in^2
y2 = t+0.5*(d-t);## inches
y_bar = (a1*y1+a2*y2)/(a1+a2);## inches
I_AB = (1/3.)*b*t**3. + (1./12.)*t*(d-t)**3 + (b-2*t)*(b-t)**2.;## in^4
I_xx = I_AB - (a1+a2)*y_bar**2.;## in^4
q = (F/(b*I_xx))*b*t*(y_bar-0.5*t);## ton/in^2
q_max = (F/(t*I_xx))*(b*t*(y_bar-0.5*t) + 0.5*t*(y_bar-t)*(y_bar-t));## tons/in^2
print'%s %.2f %s'%('The maximum intensity of shear stress at the N.A is q_max = ',q_max,' tons/in^2');

The maximum intensity of shear stress at the N.A is q_max =  4.26  tons/in^2


## Ex25-pg190¶

In [2]:
#calculate tensile and compressive various inches at 2,4, 6 and at N.A points
import math
def func(p,q):
p1 = 0.5*p + math.sqrt(q**2. + 0.25*p**2);
p2 = 0.5*p - math.sqrt(q**2. + 0.25*p**2);
theta = 0.5*math.atan(2.*q/p) * 180/math.pi;
return p1,p2,theta
b = 5.;## inches
d = 12.;## inches
F = 4800. ;## lb-wt
M = 192000.;## lb-inches
I = (1/12.)*b*d**3.;## in**4

##At 6 inches above the N.A
p6 = M*6./I ; ## lb/in**2
q6 = 0;
[p1_6,p2_6,theta6] = func(p6,q6);

##At 4 inches above the N.A
p4 = M*4/I;## lb/in**2
q4 = (F/(I*b))*b*(0.5*d-4)*b;
[p1_4,p2_4,theta4] = func(p4,q4);

##At 2 inches above the N.A
p2 = M*2./I;## lb/in**2
q2 = (F/(I*b))*b*(0.5*d-2)*4.;
[p1_2,p2_2,theta2] = func(p2,q2);

##At the N.A
p = 0.;##
q = F*b*0.5**3*d**2./(I*b);## lb/in**2
p1 = q;## lb/in**2
p2 = -q;## lb/in**2

print'%s %.2f %s  %.2f %s '%('At 6 inches above the N.A, p1 =',p1_6,' lb/in**2.'and  'compressive, and p2 =  ',p2_6,'');
print'%s %.2f %s  %.2f %s  %.2f %s  %.2f %s '%(' At 4 inches above the N.A, p1 =',p1_4,' lb/in**2.'and ' compressive, and p2 = ',-p2_4,' lb/in**2 'and'tensile theta1 = ',theta4,' degrees'  'theta2 = ',theta4+90,' degrees');
print'%s %.2f %s  %.2f %s  %.2f %s  %.2f %s'%(' At 2 inches above the N.A, p1 = ',p1_2,' lb/in**2.'and' compressive' and 'p2 = ',-p2_2,' lb/in**2.,'and 'tensile theta1 = ',theta2,' degrees 'and ' theta2 = ',theta2+90,' degrees');
print'%s %.2f %s  %.2f %s'%(' At the N.A, p1 = ',p1,' lb/in**2.,'and ' compressive, and p2 = ',-p2,'.,tensile ');

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

At 6 inches above the N.A, p1 = 1600.00 compressive, and p2 =    0.00
At 4 inches above the N.A, p1 = 1070.82  compressive, and p2 =   4.15 tensile theta1 =   3.56  degreestheta2 =   93.56  degrees
At 2 inches above the N.A, p1 =  553.88 p2 =   20.54 tensile theta1 =   10.90  theta2 =   100.90  degrees
At the N.A, p1 =  120.00  compressive, and p2 =   120.00 .,tensile


## Ex26-pg194¶

In [2]:
import math
#calculate At the top, principal stresses and   The principal stresse across the diagonal and compressive on one plane
b = 10.;## inches
d = 8.;## inches
t1 = 1.;## inch
t2 = 0.6;## inch
M = 500.;## ton-inches
F = 25.;## tons
I = (1./12.)*(d*b**3. - (d-t2)*d**3.);## in**4

##At the top
p = M*b/(2.*I);## tons/in**2
q = 0.;
p1 = p;## tons/in**2
p2 = 0.;
print'%s %.1f %s %.1f %s'%('At the top, principal stresses are  p1 =',p1,'tons/in**2'' p2 = ',p2,' tons/in**2')

##In the web, 4 inches from the N.A
p = M*d/(2.*I);## tons/in**2
q = F*d*t1*0.5*(d+t1)/(I*t2);## tons/in**2
theta = 0.5*math.atan(2*q/p);
theta1 = theta*180/math.pi;
theta2 = theta1+90;
p1 = 0.5*p + math.sqrt(q**2 + 0.25*p**2);## tons/in**2
p2 = 0.5*p - math.sqrt(q**2 + 0.25*p**2);## tons/in**2
print'%s %.3f %s %.1f %s %.1f %s %.1f %s '%(' In the web, 4 inches from the N.A.:'' The principal stresse are p1 = ',p1,' tons/in**2.,compressive\n p2 = ',-p2,' tons/in**2.,tensile\n theta1 =',theta1,' degrees\n theta2 = ',theta2,' degrees')

##At the N.A
p = 0.;
q = (F/(I*t2))*(d*t1*0.5*(d+t1) + t2*0.5*d*2*t1);
p1 = q;## tons/in**2
p2 = -q;##tons/in**2
print'%s %.1f %s %.1f %s'%(' The principal stresse across the diagonal are ',q,' tons/in**2., compressive on one plane and',q,' tons/in**2., tensile on the other.');

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

At the top, principal stresses are  p1 = 7.1 tons/in**2 p2 =  0.0  tons/in**2
In the web, 4 inches from the N.A.: The principal stresse are p1 =  7.987  tons/in**2.,compressive
p2 =  2.3  tons/in**2.,tensile
theta1 = 28.2  degrees
theta2 =  118.2  degrees
The principal stresse across the diagonal are  4.8  tons/in**2., compressive on one plane and 4.8  tons/in**2., tensile on the other.


## Ex27-pg195¶

In [3]:
import math
#calculate maximum intensity of shear stress
W = 10.;## tons
l = 16.;## feet
f = 15/2.;## tons/in^2
##section modulus required
SM = W*l*12./(8.*f);## in^3
##for this section modulus
l1 = 12.;## inches
b1 = 5.;## inches
t1 = 0.55;## inches
t2 = 0.35;## inches
I_xx = 220.;## in^4
F_max = 5.;## tons
q_max = (F_max/(I_xx*t2))*(F_max*t1*(0.5*l1-0.5*t1) + t2*0.5*(0.5*l1-t1)**2);## tons/in^2
print'%s %.2f %s'%('The maximum intensity of shear stress is q_max = ',q_max,' tons/in^2');

The maximum intensity of shear stress is q_max =  1.36  tons/in^2


## Ex28-pg198¶

In [4]:
import math
#calculate Skin stresse in steel plate  and The total moment of resistance
b = 9/2.;## inches
D = 12.;## inches
d = 10.;## inches
t = 1/2.;## inches
f_w = 1000.;## lb/in**2
m = 18.;##m = E_s/E_w
f_t = m*d*f_w/D ;## lb/in**2
M_w = f_w*(1./6.)*2.*b*D**2.;## lb-inches
M_s = f_t*(1/6.)*t*d**2;## lb-inches
M = M_w + M_s;## lb-inches
print'%s %.1f %s %.1f %s'%('Skin stresse in steel plate is, M_s = ',M_s,' lb-inches'' The total moment of resistance is M =',M,' lb-inches')

Skin stresse in steel plate is, M_s =  125000.0  lb-inches The total moment of resistance is M = 341000.0  lb-inches