# Chapter 6 : Analysis of structures¶

## Example 6.1 Page No : 294¶

In :
#Entire truss
#Applying sum(M_C) = 0

# Given Data
E = (10.*12+5*6)/3;			#kN
#Applying sum Fx = 0
Cx = 0.
#Applying sumFy = 0
Cy = 10.+5-E;			#kN

# Calculations
#At joint A
#By proportion 10kN/4 = F_AB/3 = F_AD/5
F_AB = 10./4*3;			#kN, force in member AB
F_DA = 10./4*5;			#kN, force in member AD

#At joint D
F_DB = F_DA;			#kN, force in member DB
F_DE = 2*3./5*F_DA;			#kN, force in member DE

#At joint B
#applying sumFy = 0
F_BE = 5./4*(-5-4./5*F_DB);			#kN, force in member BE
#Applying sumFx = 0
F_BC = F_AB+3./5*F_DB-3./5*F_BE;			#kN, force in member BC
#At joint E
#Applying sumFx = 0
F_EC = -5./3*(F_DE-3./5*F_BE);			#kN, Force in member EC

# Results
print "The forces in member of truss are  F_AB =  %.1f kN T  \
\nF_AD =  %.1f kN C,  \
\nF_DB =  %.1f kN T  \
\nF_DE =  %.0f kN C  \
\nF_BE =  %.2f kN   \
\nF_BC =  %.2f kN  \
\nF_EC =  %.2f kN "%(F_AB,F_DA,F_DB,F_DE,F_BE,F_BC,F_EC);
#Variation in answe because of round off

The forces in member of truss are  F_AB =  7.5 kN T
F_DB =  12.5 kN T
F_DE =  15 kN C
F_BE =  -18.75 kN
F_BC =  26.25 kN
F_EC =  -43.75 kN


## Example 6.2 Page No : 306¶

In :
# Given Data
#Entire truss
v1 = 140.;			#kn, verical force 1
v2 = 140.;			#kN, Vertical force 2
h = 80.;			#kN , Horizontal force
#Applying sum(M_B) = 0
J = (v1*4+v2*12+h*5)/16;			#kN

# Calculations and Results
#Applying sum Fx = 0
Bx = -h;			#kN, negative sign shows it is along negative x axis

#Applying sumFy = 0

By = v1+v2-J;			#kN

#Force in member EF
#Applying sumFy = 0
F_EF = By-v2;			#kN, Force in member EF
print "Force in member EF is %.0f kN  Negative sign shows member is in compression "%(F_EF);

#Force in member GI
F_GI = (-J*4-Bx*5)/5;			#kN Force in member GI
print "Force in member GI is %.0f kN  Negative sign shows member is in compression "%(F_GI);
#Answer difference is because of rounding off variables

Force in member EF is -25 kN  Negative sign shows member is in compression
Force in member GI is -52 kN  Negative sign shows member is in compression


## Example 6.3 Page No : 307¶

In :
import math

# Given Data
#Entire truss
vB = 1.;			#kN, verical force at B
vD = 1.;			#kN, verical force at D
vF = 1.;			#kN, verical force at F
vH = 1.;			#kN, verical force at H
vJ = 1.;			#kN, verical force at J
vC = 5.;			#kN, verical force at C
vE = 5.;			#kN, verical force at E
vG = 5.;			#kN, verical force at G
h = 8.;			#m, height
v = 5.;			#m, horizontal dismath.tance between successive node

A = 12.50;			#kN, reaction at A
L = 7.50;			#kN, reaction at L

# Calculations and Results
alpha = math.atan(h/3./v);			# rad, angle made by inclined members with  X axis
#alpha = alpha/math.pi*180;			# Conversion of angle into degrees

#Force in member GI
#Applying sum(M_H) = 0
F_GI = (L*2*v-vJ*v)/(2*v*math.tan(alpha));			#kN Force in member GI
print "Force in member GI is %.2f kN  "%(F_GI);

#Force in member FH
#Applying sum(M_G) = 0
F_FH = (L*3*v-vH*v-vJ*2*v)/(-h*math.cos(alpha));			#kN, Force in member FH
print "Force in member FH is %.2f kN  Negative sign shows member is in compression "%(F_FH);

#Force in member GH
be = math.atan(v/(2*v*math.tan(alpha)));			#rad, as math.tan(be) = GI/HI
#Applying sum(M_L) = 0
F_GH = (-vH*v-vJ*2*v)/(3*v*math.cos(be));			#kN, Force in member FH
print "Force in member GH is %.3f kN  Negative sign shows member is in compression "%(F_GH);

Force in member GI is 13.12 kN
Force in member FH is -13.81 kN  Negative sign shows member is in compression
Force in member GH is -1.371 kN  Negative sign shows member is in compression


## Example 6.4 Page No : 319¶

In :
import math
#Entire truss
#Applying sum(Fy) = 0

# Given Data
Ay = 480.;			#N, Y component of reaction at A
#Applying sum(M_A) = 0
B = 480*100./160;			#N,  reaction at B
#Applying sum(Fx) = 0
Ax = -300.;			#N, X component of reaction at A

# Calculations and Results
#Free body member BCD
#Applying sum(M_C) = 0
F_DE = (-480*100.-B*60)/(math.sin(alpha)*250);			#N,  Force in link DE
print "Force in link DE is F_DE = %.0f N Negative sign shows force is compressive"%(F_DE);
#Applying sum(Fx) = 0
Cx = F_DE*math.cos(alpha)-B;			#N, X component of force exerted at C
#Applying sum(Fy) = 0
Cy = F_DE*math.sin(alpha)+Ay;			#N, Y component of force exerted at C
print "Components of force exerted at C is Cx = %.0f N and Cy = %.0f N "%(Cx,Cy);

Force in link DE is F_DE = -561 N Negative sign shows force is compressive
Components of force exerted at C is Cx = -795 N and Cy = 216 N


## Example 6.5 Page No : 320¶

In :
# Given Data
P = 18.;			#kN, Force applied at D
AF = 3.6;			#m, Length AF
EF = 2.;			#m, Length EF
ED = 2.;			#m, Length ED
DC = 2.;			#m, Length DC
#Entire frame

# Calculations and Results
#Applying sum(M_F) = 0
Ay = -P*(EF+ED)/AF;			#kN, Y component of reaction at A

#Applying sum(Fx) = 0
Ax = -P;			#kN, X component of reaction at A
#Applying sum(Fy) = 0
F = -Ay;			#kN,  reaction at B

print "Components of force exerted at A is Ax = %.0f kN and Ay = %.0f kN "%(Ax,Ay);
print "Force exerted at F is F = %.0f kN "%(F);
#Free body member BE
#Applying sum(Fx) = 0
#B = E, and as it is 2 force member
By = 0;
Ey = 0;

#Member ABC
#Applying sum(Fy) = 0
Cy = -Ay;			#kN, Y component of force exerted at C
#Applying sum(M_C) = 0
B = (Ay*AF-Ax*(DC+ED+EF))/(ED+DC);			#kN,  Force in link DE
print "Force exerted at B is B = %.0f kN  "%(B);
#Applying sum(Fx) = 0
Cx = -Ax-B;			#kN, X component of force exerted at C

print "Components of force exerted at C is Cx = %.0f kN and Cy = %.0f kN "%(Cx,Cy);
print "Negative signs shows forces are in negative direction"

Components of force exerted at A is Ax = -18 kN and Ay = -20 kN
Force exerted at F is F = 20 kN
Force exerted at B is B = 9 kN
Components of force exerted at C is Cx = 9 kN and Cy = 20 kN
Negative signs shows forces are in negative direction


## Example 6.6 Page No : 320¶

In :
import math
from numpy.linalg import solve

# Given Data
P = 3.;			#kN, Horizontal Force applied at A
AB = 1.;			#m, perpendicular dismath.tance between A and B
BD = 1.;			#m, perpendicular dismath.tance between D and B
CD = 1.;			#m, perpendicular dismath.tance between C and D
FC = 1.;			#m, perpendicular dismath.tance between C and F
EF = 2.4;			#m, perpendicular dismath.tance between E and F
#Entire frame

# Calculations
#Applying sum(M_E) = 0
Fy = P*(AB+BD+CD+FC)/EF;			#kN, Y component of reaction at F

#Applying sum(Fy) = 0
Ey = -Fy;			#kN, Y component of  reaction at E

#Free body member ACE
#Applying sum(Fy) = 0, and sum(M_E) = 0 we get 2 equation
A = [[-AB/math.sqrt(AB**2+EF**2) ,CD/math.sqrt(CD**2+EF**2) ] , [ -EF/math.sqrt(AB**2+EF**2)*(AB+BD+CD+FC) , -EF/math.sqrt(CD**2+EF**2)]];			# Matrix of coefficients
B = [[Ey],[-P*(AB+BD+CD+FC)]];			# Matrix B
X = solve(A,B);			#kN Solution matrix
F_AB = X;			#kN, Forec inmember AB
F_CD = X;			#kN, Forec inmember CD
Ex = -P-EF/math.sqrt(AB**2+EF**2)*F_AB-EF/math.sqrt(CD**2+EF**2)*F_CD;			#kN, X component of force exerted at E
#Free body : Entire frame
#Applying sum(F_X) = 0
Fx = -P-Ex;			#kN, X component of force exetered at F
print "Components of force exerted at F is Fx = %.1f kN and Fy = %.0f kN "%(Fx,Fy);
print "Force in member AB is F_AB = %.1f kN "%(F_AB);
print "Force in member CD is F_CD = %.1f kN "%(F_CD);
print "Components of force exerted at E is Ex = %.1f kN and Ey = %.1f kN "%(Ex,Ey);

print "Negative signs shows forces are in negative direction"

Components of force exerted at F is Fx = -2.4 kN and Fy = 5 kN
Force in member AB is F_AB = 5.2 kN
Force in member CD is F_CD = -7.8 kN
Components of force exerted at E is Ex = -0.6 kN and Ey = -5.0 kN
Negative signs shows forces are in negative direction