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-rw-r--r--3772/CH10/EX10.1/Ex10_1.sce34
-rw-r--r--3772/CH10/EX10.2/Ex10_2.sce57
-rw-r--r--3772/CH10/EX10.3/Ex10_3.sce65
-rw-r--r--3772/CH10/EX10.4/Ex10_4.sce42
-rw-r--r--3772/CH10/EX10.5/Ex10_5.sce62
-rw-r--r--3772/CH10/EX10.6/Ex10_6.sce51
-rw-r--r--3772/CH10/EX10.7/Ex10_7.sce53
-rw-r--r--3772/CH10/EX10.8/Ex10_8.sce53
8 files changed, 417 insertions, 0 deletions
diff --git a/3772/CH10/EX10.1/Ex10_1.sce b/3772/CH10/EX10.1/Ex10_1.sce
new file mode 100644
index 000000000..1cc6b9416
--- /dev/null
+++ b/3772/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,34 @@
+// Problem no 10.1,Page No.249
+
+clc;clear;
+close;
+
+//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
+printf("The Forces in Each members are as follows:F_AB = %.f kN",F_AB)
+printf("\n :F_AC = %.f kN",F_AC)
+printf("\n :F_DB %.2f",F_DB);printf(" KN(compression)")
+printf("\n :F_BC %.2f",F_BC);printf(" KN(Tension)")
+printf("\n :F_CE %.2f",F_CE);printf(" KN(Tension)")
+printf("\n :F_DC %.2f",F_DC);printf(" KN (compression)" )
diff --git a/3772/CH10/EX10.2/Ex10_2.sce b/3772/CH10/EX10.2/Ex10_2.sce
new file mode 100644
index 000000000..91d6396ba
--- /dev/null
+++ b/3772/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,57 @@
+// Problem no 10.2,Page No.250
+
+clc;clear;
+close;
+
+//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
+F_AD=200-F_AC*sin(45*%pi*180**-1)
+
+//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
+printf("The Forces in Each members are as follows:F_AB = %.1f kN",F_AB)
+printf("\n :F_CB = %.1f kN (compressive)",F_CB)
+printf("\n :F_AC %.2f",F_AC);printf(" KN(Tensile)")
+printf("\n :F_AD=%.1f kN (Tensile)",F_AD)
+printf("\n :F_EC=%.1f kN N(Compressive)",F_EC)
+printf("\n :F_DC=%.1f kN N(Compressive)",F_DC)
+printf("\n :F_DE %.2f",F_DE);printf(" KN(Tensile)")
+printf("\n :F_DF = %.f kN",F_DF)
+printf("\n :F_EF = %.f kN",F_EF)
diff --git a/3772/CH10/EX10.3/Ex10_3.sce b/3772/CH10/EX10.3/Ex10_3.sce
new file mode 100644
index 000000000..23777b820
--- /dev/null
+++ b/3772/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,65 @@
+// Problem no 10.3,Page No.252
+
+clc;clear;
+close;
+
+//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
+printf("Forces in Each members are as follows:F_GD %.1f kN (compression)",F_GD)
+printf("\n :F_DC %.2f",F_DC);printf(" KN(Tension)" )
+printf("\n :F_GC %.1f kN (Tension)",F_GC)
+printf("\n :F_FG %.1f kN (Compression)",F_FG)
+printf("\n :F_FC %.1f kN(compression)",F_FC)
+printf("\n :F_CB %.2f",F_CB);printf(" KN(Tension)")
+printf("\n :F_FB %.1f kN(compression)",F_FB)
+printf("\n :F_EF %.2f",F_EF);printf(" KN(compression)")
+printf("\n :F_EB %.2f",F_EB);printf(" KN(Tension)")
+printf("\n :F_BA %.2f",F_BA);printf(" KN(Tension)")
+printf("\n :F_AE %.2f",F_AE);printf(" KN(compression)")
diff --git a/3772/CH10/EX10.4/Ex10_4.sce b/3772/CH10/EX10.4/Ex10_4.sce
new file mode 100644
index 000000000..a07330b75
--- /dev/null
+++ b/3772/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,42 @@
+// Problem no 10.4,Page No.253
+
+clc;clear;
+close;
+
+//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
+printf("Forces in Each members are as follows:F_1 %.2f",F_1);printf(" KN(Tension)")
+printf("\n :F_5 %.2f",F_5);printf(" KN(compression)")
+printf("\n :F_6 %.2f",F_6);printf(" KN(compression)")
+printf("\n :F_2 %.2f",F_2);printf(" KN(Tension)")
+printf("\n :F_7 %.2f",F_7);printf(" KN(Tension)")
+printf("\n :F_4 %.2f",F_4);printf(" KN(compression)")
+printf("\n :F_3 %.2f",F_3);printf(" KN(compression)")
+printf("\n :F_8 %.2f",F_8);printf(" KN(Tension)")
diff --git a/3772/CH10/EX10.5/Ex10_5.sce b/3772/CH10/EX10.5/Ex10_5.sce
new file mode 100644
index 000000000..af85fdd5a
--- /dev/null
+++ b/3772/CH10/EX10.5/Ex10_5.sce
@@ -0,0 +1,62 @@
+// Problem no 10.5,Page No.256
+
+clc;clear;
+close;
+
+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
+printf("Forces in Each members are as follows:F_BD %.2f",F_BD);printf(" KN(compression)")
+printf("\n :F_BE %.2f",F_BE);printf(" KN(Tension)")
+printf("\n :F_DE %.2f",F_DE);printf(" KN(compression)")
+printf("\n :F_CD %.2f",F_CD);printf(" KN(compression)")
+printf("\n :F_CE %.2f",F_CE);printf(" KN(Tension)")
+printf("\n :F_AC %.2f",F_AC);printf(" KN(compression)")
+printf("\n :F_AE %.2f",F_AE);printf(" KN(Tension)")
diff --git a/3772/CH10/EX10.6/Ex10_6.sce b/3772/CH10/EX10.6/Ex10_6.sce
new file mode 100644
index 000000000..c86506905
--- /dev/null
+++ b/3772/CH10/EX10.6/Ex10_6.sce
@@ -0,0 +1,51 @@
+// Problem no 10.6,Page No.258
+
+clc;clear;
+close;
+
+//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
+F_AD=2000*2**0.5*3**0.5*2**-1-1000*2**0.5*3**-1 //N
+
+//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=atan(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
+F_DB=(F_AD-1000*2**0.5)*(cos(60*%pi*180**-1))**-1
+
+//Result
+printf("The Pull in chain is %.2f",P);printf(" N")
+printf("\n Force in the each members are as follows:F_BA %.2f",F_BA);printf(" KN(compressive)")
+printf("\n :F_AD %.2f",F_AD);printf(" KN(Tensile)")
+printf("\n :F_CB %.2f",F_CB);printf(" KN(compression)")
+printf("\n :F_CD %.2f",F_CD);printf(" KN(Tensile)")
+printf("\n :F_DB %.2f",F_DB);printf(" KN(compressive)")
diff --git a/3772/CH10/EX10.7/Ex10_7.sce b/3772/CH10/EX10.7/Ex10_7.sce
new file mode 100644
index 000000000..7c909293b
--- /dev/null
+++ b/3772/CH10/EX10.7/Ex10_7.sce
@@ -0,0 +1,53 @@
+// Problem no 10.7,Page No.261
+clc;
+clear;
+close;
+
+
+//Calculations
+
+theta=atan(1*2**-1)*(180*%pi**-1) //Radian
+
+//Taking moment about A
+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
+printf("Forces in Each members are as follows:F_AB %.2f",F_AB);printf(" KN(Tensile)")
+printf("\n :F_DC %.2f",F_DC);printf(" KN(compression)")
+printf("\n :F_CB %.2f",F_CB);printf(" KN(Tensile)")
+printf("\n :F_DB %.2f",F_DB);printf(" KN(Tensile)")
+printf("\n :F_DE %.2f",F_DE);printf(" KN(compression)")
+printf("\n :F_EB %.2f",F_EB);printf(" KN")
diff --git a/3772/CH10/EX10.8/Ex10_8.sce b/3772/CH10/EX10.8/Ex10_8.sce
new file mode 100644
index 000000000..9c7f6088a
--- /dev/null
+++ b/3772/CH10/EX10.8/Ex10_8.sce
@@ -0,0 +1,53 @@
+// Problem no 10.8,Page No.262
+
+clc;clear;
+close;
+
+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
+L_AD=3.6;L_BE=3.6 //m //Length of AD & BE
+
+//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
+sigma_1=(F_d*L_DE+F_c*(L_CD+L_DE)-R_A*L_DE)*L_AD**-1 //Force i member DE
+
+//Taking moment @ D
+sigma_3=(F_c*L_CD)*L_AD**-1 //KN //force in member AB
+
+
+//Consider triangle DBE
+theta=atan(L_BE*L_DE**-1)*(180*%pi**-1)
+
+//Taking moment @ A
+sigma_2=(-sigma_1*L_AD+F_c*L_CD)*(L_AD*cos(theta*%pi*180**-1))**-1 //Force in member F_DE
+
+//Now considering section 2-2 passing through members AB,AD,CD and taking left hand F.B.D
+
+//Taking moment @C
+sigma_5=(R_A*L_CD-sigma_3*L_AD)*L_CD**-1 //Force in member AD
+
+//Taking moment @A=0
+sigma_4=F_c*L_CD*L_AD**-1 //Force in member CD
+
+
+//Result
+printf("Force in member CD is %.2f",sigma_4);printf(" KN(Compressive)")
+printf("\n Force in member AD is %.2f",sigma_5);printf(" KN(Tensile)")
+printf("\n Force in member BD is %.2f",sigma_2);printf(" KN(Compression)")
+printf("\n Force in member AB is %.2f",sigma_1);printf(" KN(Tension)")
+
+// Answer is wrong in the textbook.
generated by cgit (git 2.34.1) at 2025-03-11 06:27:18 +0000