# Chapter 10:Analysis Of Framed Structures¶

## Problem no 10.1,Page No.249¶

In :
import math

#Initilization of variables

#Consider Equilibrium of joint A
#As there are no Load applied at A members AC and AB have nothing to Balance
#So they are null members
F_AB=0
F_AC=0

#Consider Equilibrium of joint B

#Applying the summation of horizontal forces  we get
F_DB=4*(cos(45*pi*180**-1))**-1

#Applying the summation of vertical forces  we get
F_BC=F_DB*sin(45*pi*180**-1)

#Consider Equilibrium of joint B

#Applying the summation of vertical forces  we get
F_CE=4*(sin(45*pi*180**-1))**-1

#Applying the summation of horizontal forces  we get
F_DC=F_CE*cos(45*pi*180**-1)

#Result
print"The Forces in Each members are as follows:F_AB",F_AB,"KN"
print"                                         :F_AC",F_AC,"KN"
print"                                         :F_DB",round(F_DB,2),"KN(compression)"
print"                                         :F_BC",round(F_BC,2),"KN(Tension)"
print"                                         :F_CE",round(F_CE,2),"KN(Tension)"
print"                                         :F_DC",round(F_DC,2),"KN (compression)"

The Forces in Each members are as follows:F_AB 0 KN
:F_AC 0 KN
:F_DB 5.66 KN(compression)
:F_BC 4.0 KN(Tension)
:F_CE 5.66 KN(Tension)
:F_DC 4.0 KN (compression)


## Problem no 10.2,Page No.250¶

In :
import math

#Initilization of variables

#Taking moment at Pt A we get
R_B=100*8*4**-1

#Applying the summation of vertical forces  we get
R_AV=-R_B

#Applying the summation of horizontal forces  we get
R_H=100

#joint B

#Applying the summation of vertical forces  we get
F_CB=R_B

#Applying the summation of horizontal forces  we get
F_AB=0 #As there is no force to balance in horizontal direction

#joint A

#Applying the summation of horizontal forces  we get
F_AC=R_H*(cos(45*pi*180**-1))**-1

#Applying the summation of vertical forces  we get

#joint C

#Applying the summation of vertical forces  we get
F_EC=200-F_AC*cos(45*pi*180**-1)

#Applying the summation of horizontal  forces  we get
F_DC=F_AC*cos(45*pi*180**-1)

#joint D

#Applying the summation of horizontal  forces  we get
F_DE=F_DC*(cos(45*pi*180**-1))**-1

#DF and EF are null members at this joint as each member individually has nothing to balance
F_DF=0
F_EF=0

#Result
print"The Forces in Each members are as follows:F_AB",F_AB,"KN"
print"                                         :F_CB",F_CB,"KN(compressive)"
print"                                         :F_AC",round(F_AC,2),"KN(Tensile)"
print"                                         :F_EC",F_EC,"KN(Compressive)"
print"                                         :F_DC",F_DC,"KN(compressive)"
print"                                         :F_DE",round(F_DE,2),"KN(Tensile)"
print"                                         :F_DF",F_DF,"KN"
print"                                         :F_EF",F_EF,"KN"

The Forces in Each members are as follows:F_AB 0 KN
:F_CB 200.0 KN(compressive)
:F_AC 141.42 KN(Tensile)
:F_EC 100.0 KN(Compressive)
:F_DC 100.0 KN(compressive)
:F_DE 141.42 KN(Tensile)
:F_DF 0 KN
:F_EF 0 KN


## Problem no 10.3,Page No.252¶

In :
import math

#Initilization of variables

#taking moment at pt A we get
R_D=(90*6+120*3)*9**-1 #Reaction at Pt D

#Joint D

#Applying the summation of vertical forces  we get
F_GD=100*(sin(60*pi*180**-1))**-1

#Applying the summation of horizontal  forces  we get
F_DC=F_GD*cos(60*pi*180**-1)

#Joint G

#Applying the summation of vertical forces  we get
F_GC=F_GD

#Applying the summation of horizontal  forces  we get
F_FG=F_GD*cos(60*pi*180**-1)+F_GC*cos(60*pi*180**-1)

#joint C

#Applying the summation of vertical forces  we get
F_FC=(115.5*sin(60*pi*180**-1)-90)*(sin(60*pi*180**-1))**-1

#Applying the summation of horizontal  forces  we get
F_CB=F_DC+F_GC*cos(60*pi*180**-1)+F_FC*cos(60*pi*180**-1)

#joint F

#Applying the summation of vertical forces  we get
F_FB=F_FC

#Applying the summation of horizontal  forces  we get
F_EF=F_FG+F_FC*cos(60*pi*180**-1)+F_FB*cos(60*pi*180**-1)

#Joint B

#Applying the summation of vertical forces  we get
F_EB=(120-F_FB*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1

#Applying the summation of horizontal  forces  we get
F_BA=F_CB+F_FB*cos(60*pi*180**-1)-F_EB*cos(60*pi*180**-1)

#Joint E

#Applying the summation of vertical forces  we get
F_AE=F_EB

#Result
print"Forces in Each members are as follows:F_GD",round(F_GD,1),"KN(compression)"
print"                                     :F_DC",round(F_DC,2),"KN(Tension)"
print"                                     :F_GC",round(F_GC,1),"KN(Tension)"
print"                                     :F_FG",round(F_FG,1),"KN(compression)"
print"                                     :F_FC",round(F_FC,1),"KN(compression)"
print"                                     :F_CB",round(F_CB,2),"KN(Tension)"
print"                                     :F_FB",round(F_FB,1),"KN(Tension)"
print"                                     :F_EF",round(F_EF,2),"KN(compression)"
print"                                     :F_EB",round(F_EB,2),"KN(Tension)"
print"                                     :F_BA",round(F_BA,2),"KN(Tension)"
print"                                     :F_AE",round(F_AE,2),"KN(compression)"

Forces in Each members are as follows:F_GD 115.5 KN(compression)
:F_DC 57.74 KN(Tension)
:F_GC 115.5 KN(Tension)
:F_FG 115.5 KN(compression)
:F_FC 11.6 KN(compression)
:F_CB 121.26 KN(Tension)
:F_FB 11.6 KN(Tension)
:F_EF 127.05 KN(compression)
:F_EB 126.99 KN(Tension)
:F_BA 63.55 KN(Tension)
:F_AE 126.99 KN(compression)


## Problem no 10.4,Page No.253¶

In :
import math

#Initilization of variables

#JOint D

#Applying the summation of vertical forces  we get
F_1=6*sin(30*pi*180**-1)**-1

#Applying the summation of horizontal  forces  we get
F_5=F_1*cos(30*pi*180**-1)

#Joint C

#Resolving forces perpendicular to plane
F_6=10*cos(30*pi*180**-1)

#Resolving forces parallel to plane
F_2=F_1+10*cos(60*pi*180**-1)

#Joint E

#Applying the summation of vertical forces  we get
F_7=(8+F_6*sin(60*pi*180**-1))*(sin(60*pi*180**-1))**-1
F_4=F_5+F_6*cos(60*pi*180**-1)+F_7*cos(60*180**-1*pi)

#Resolving forces perpendicular to plane
F_3=F_7*sin(60*pi*180**-1)

#Resolving forces parallel to plane
F_8=F_2+F_7*cos(30*pi*180**-1)

#Result
print"Forces in Each members are as follows:F_1",round(F_1,2),"KN(Tension)"
print"                                     :F_5",round(F_5,2),"KN(compression)"
print"                                     :F_6",round(F_6,2),"KN(compression)"
print"                                     :F_2",round(F_2,2),"KN(Tension)"
print"                                     :F_7",round(F_7,2),"KN(Tension)"
print"                                     :F_4",round(F_4,2),"KN(compression)"
print"                                     :F_3",round(F_3,2),"KN(compression)"
print"                                     :F_8",round(F_8,2),"KN(Tension)"

Forces in Each members are as follows:F_1 12.0 KN(Tension)
:F_5 10.39 KN(compression)
:F_6 8.66 KN(compression)
:F_2 17.0 KN(Tension)
:F_7 17.9 KN(Tension)
:F_4 23.67 KN(compression)
:F_3 15.5 KN(compression)
:F_8 32.5 KN(Tension)


## Problem no 10.5,Page No.256¶

In :
import math

#Initilization of variables

BC=6 #m

#Calculations

AB=2*BC*(3**0.5)**-1

#Taking moment about B we get
R_A=-(-2000*3-1000*6)*(12*(3**0.5)**-1)**-1 #reaction at the roller support A

#The resultant of all the three Loads is 4000 N acting at right angle to BC at D

#Resolving it vertically we have
V=4000*sin(60*pi*180**-1)

#Resolving it horizontal we have
H=4000*cos(60*pi*180**-1)

#Applying the summation of vertical forces  we get
R_B_v=V-R_A

#Applying the summation of horizontal  forces  we get
R_B_h=H
R_B=((R_B_v)**2+(R_B_h)**2)**0.5

tan_theta=R_B_v*R_B_h**-1

#Joint B

#Applying the summation of vertical forces  we get
F_BD=1000*(3**0.5)*2

#Applying the summation of horizontal  forces  we get
F_BE=R_B_h+F_BD*cos(30*pi*180**-1)

#Joint D
F_DE=2000 #N
F_CD=F_BD

#Consider equilibrium of truss to the Left of section 2-2
F_CE=R_A*AB*(sin(30*pi*180**-1)*6)**-1

#Joint A

#Applying the summation of vertical forces  we get
F_AC=R_A*(sin(60*pi*180**-1))**-1

#Applying the summation of horizontal  forces  we get
F_AE=F_AC*cos(60*pi*180**-1)

#Result
print"Forces in Each members are as follows:F_BD",round(F_BD,2),"KN(compression)"
print"                                     :F_BE",round(F_BE,2),"KN(Tension)"
print"                                     :F_DE",round(F_DE,2),"KN(compression)"
print"                                     :F_CD",round(F_CD,2),"KN(compression)"
print"                                     :F_CE",round(F_CE,2),"KN(Tension)"
print"                                     :F_AC",round(F_AC,2),"KN(compression)"
print"                                     :F_AE",round(F_AE,2),"KN(Tension)"

Forces in Each members are as follows:F_BD 3464.1 KN(compression)
:F_BE 5000.0 KN(Tension)
:F_DE 2000.0 KN(compression)
:F_CD 3464.1 KN(compression)
:F_CE 4000.0 KN(Tension)
:F_AC 2000.0 KN(compression)
:F_AE 1000.0 KN(Tension)


## Problem no 10.6,Page No.258¶

In :
import math

#Calculations

#Taking moment of the Forces about the hinge A
P=1000*2**0.5*1.2*(0.9)**-1

#Let R_AH be the Horizontal component of the reaction at A
R_AH=P-1000*2**0.5
R_A=((R_AH)**2+(1000*2**0.5)**2)**0.5

#Resolving the forces vertically we get
R_AV=1000*2**0.5 #vertical component of the reaction at A

#joint A

#Resolving vertically we get
F_BA=1000*2**0.5*(sin(30*pi*180**-1))**-1

#Resolving horizontally we get

#Joint C

BD=1.2*sin(30*pi*180**-1)
BE=0.6*sin(30*pi*180**-1)
ED=0.6*cos(30*pi*180**-1)
CE=0.9-0.52

theta=arctan(BE*CE**-1)*(180*pi**-1)

F_CB=P*(sin(38.29*pi*180**-1))**-1

#Resolving vertically
F_CD=F_CB*cos(theta*pi*180**-1)

#Joint D

#Resolving horizontally

#Result
print"The Pull in chain is",round(P,2),"N"
print"Force in the each members are as follows:F_BA",round(F_BA,2),"KN(compressive)"
print"                                        :F_CB",round(F_CB,2),"KN(compression)"
print"                                        :F_CD",round(F_CD,2),"KN(Tensile)"
print"                                        :F_DB",round(F_DB,2),"KN(compressive)"

The Pull in chain is 1885.62 N
Force in the each members are as follows:F_BA 2828.43 KN(compressive)
:F_CB 3043.08 KN(compression)
:F_CD 2388.46 KN(Tensile)
:F_DB 1127.74 KN(compressive)


## Problem no 10.7,Page No.261¶

In :
import math

#Calculations

R_EH=10*8*4**-1

#Horizontal component of reaction at A
R_AH=20 #KN

#Applying the summation of horizontal  forces  we get
F_AB=20*cos(theta*pi*180**-1)**-1

#Applying the summation of vertical forces  we get
R_AV=10*5**0.5*sin(theta*pi*180**-1)

#Vertical Reaction at E
R_EV=0

#Joint C

#Applying the summation of vertical forces  we get
F_DC=10*sin(theta*pi*180**-1)**-1

#Applying the summation of horizontal  forces  we get
F_CB=F_DC*cos(theta*pi*180**-1)

#Joint D

#Applying the summation of vertical forces  we get
F_DB=F_DC*sin(theta*pi*180**-1)

#Applying the summation of horizontal  forces  we get
F_DE=F_DC*cos(theta*pi*180**-1)

#Joint E

#Applying the summation of vertical forces  we get
F_EB=R_EV*sin(theta*pi*180**-1)

#Result
print"Forces in Each members are as follows:F_AB",round(F_AB,2),"KN(Tensile)"
print"                                     :F_DC",round(F_DC,2),"KN(compression)"
print"                                     :F_CB",round(F_CB,2),"KN(Tensile)"
print"                                     :F_DB",round(F_DB,2),"KN(Tensile)"
print"                                     :F_DE",round(F_DE,2),"KN(compression)"
print"                                     :F_EB",round(F_EB,2),"KN"

Forces in Each members are as follows:F_AB 22.36 KN(Tensile)
:F_DC 22.36 KN(compression)
:F_CB 20.0 KN(Tensile)
:F_DB 10.0 KN(Tensile)
:F_DE 20.0 KN(compression)
:F_EB 0.0 KN


## Problem no 10.8,Page No.262¶

In :
import math

#Initilization of variables

F_c=20 #KN #Force at C
F_d=5  #KN #Force at D
F_e=15 #KN #Force at E
F_f=10 #KN #Force at F
L_CD=3.6 #m #Length of CD
L_DE=3.6 #m #Length of DE
L_EF=4.8 #m #Length of EF

#Calculations

#Let R_A and R_B be the reactions at pts at A and B

#Taking  moment at A
R_B=-(-F_f*(L_DE+L_EF)+F_c*L_CD-F_e*L_DE)*(L_DE)**-1
R_A=50-R_B

#Considering section 1-1 through members AB,DB,DE and taking F.B.D of left side of section 1-1

#Taking moment at B

#Taking moment @ D
sigma_3=(F_c*L_CD)*L_AD**-1 #KN #force in member AB

#Consider triangle DBE
theta=arctan(L_BE*L_DE**-1)*(180*pi**-1)

#Taking moment @ A

#Now considering section 2-2 passing through members AB,AD,CD and taking left hand F.B.D

#Taking moment @C

#Taking moment @A=0

Force in member CD is 20.0 KN(Compressive)