Added new codeHEADmaster

11 files changed, 657 insertions, 0 deletions

diff --git a/3862/CH3/EX3.1/Ex3_1.sce b/3862/CH3/EX3.1/Ex3_1.sce
new file mode 100644
index 000000000..9dcac0f34
--- /dev/null
+++ b/3862/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,44 @@
+clear
+//
+
+//variable declaration
+
+//Determine the inclinations of all inclined members
+
+theta=atan(1)*180/%pi 
+
+printf("\n theta= %0.2f  °",theta)
+
+//Now at joints C, there are only two unknowns,forces in members CB and CD, say FCB and FCD.
+//Now there are two equations of equilibrium for the forces meeting at the joint and two unknown forces. Hence, the unknown forces can be determined. At joint C  sum V= 0 condition shows that the force FCB should act away from the joint C so that its vertical component balances the vertical downward load at C.
+ 
+P=40.0
+FCB=P/sin(theta*%pi/180)
+
+printf("\n FCB= %0.2f  KN",FCB)
+
+//Now sum H=0 indicates that FCD should act towards C.
+
+FCD=FCB*cos(theta*%pi/180)
+
+printf("\n FCD= %0.2f  KN",FCD)
+
+//In the present case, near the joint C, the arrows are marked on the members CB and CD to indicate forces FCB and FCD directions as found in the analysis of joint C. Then reversed directions are marked in the members CB and CD near joints B and D, respectively.
+
+FDB=40.0
+FDE=40.0
+
+printf("\n FDB= %0.2f  KN",FDB)
+
+printf("\n FDE= %0.2f  KN",FDE)
+
+//In the present case, after marking the forces in the members DB and DE, we find that analysis of joint B can be taken up.
+
+FBE=(FCB*sin(theta*%pi/180)+P)/(sin(theta*%pi/180))
+
+FBA=FCB*cos(theta*%pi/180)+FBE*cos(theta*%pi/180)
+
+printf("\n FBE= %0.2f  KN",FBE)
+printf("\n FBA= %0.2f  KN",FBA)
+//Determine the nature of forces in each member and tabulate the results. Note that if the arrow marks on a member are towards each other, then the member is in tension and if the arrow marks are away from each other, the member is in compression.
+
diff --git a/3862/CH3/EX3.10/Ex3_10.sce b/3862/CH3/EX3.10/Ex3_10.sce
new file mode 100644
index 000000000..da5656eff
--- /dev/null
+++ b/3862/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,36 @@
+clear
+//
+
+//Each load is 20 kN.
+
+//variable declaration
+
+P=20.0
+AB=18.0
+A=3.0
+
+RA=P*7/2
+RB=RA
+
+theta1=30.0*%pi/180
+a=(3*A)/(4*cos(theta1))
+//Take Section (A)–(A) and consider the equilibrium of left hand side part of the French Truss
+//Drop perpendicular CE on AB. 
+
+CE=3*A*tan(theta1)
+DE=A
+
+theta=atan(CE/DE)*180/%pi 
+printf("\n theta= %0.0f  °",theta)
+
+//moment at point A
+
+F2=(P*a*cos(theta1)*6)/(A*2*sin(theta*%pi/180))
+printf("\n F2= %0.4f KN (Tension)", F2)
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+F1=(F2*sin(theta*%pi/180)+RA-P*3)/(sin(theta1))
+printf("\n F1= %0.4f KN (Comp)",F1)
+
+F3=F1*cos(theta1)-F2*cos(theta*%pi/180)
+printf("\n F3= %0.4f KN (Tension)",F3)
diff --git a/3862/CH3/EX3.11/Ex3_11.sce b/3862/CH3/EX3.11/Ex3_11.sce
new file mode 100644
index 000000000..31e1fca85
--- /dev/null
+++ b/3862/CH3/EX3.11/Ex3_11.sce
@@ -0,0 +1,60 @@
+clear
+//
+
+//variable declaration
+
+PA=15.0       //vertical loading at point A,KN
+PB=30.0       //vertical loading at point B,KN
+PC=30.0       //vertical loading at point C,KN
+PD=30.0       //vertical loading at point D,KN
+PE=15.0       //vertical loading at point E,KN
+
+//Due to symmetry, the reactions are equal
+RA=(PA+PB+PC+PD+PE)/2
+RB=RA
+//Drop perpendicular CH on AF. 
+//in traingle ACH
+
+angleACH=45.0*%pi/180    //angleACH,°
+angleFCV=30.0*%pi/180    // FC is inclined at 30° to vertical i.e., 60° to horizontal and CH = 5 m 
+CH=5.0
+angleFCH=60.0*%pi/180
+
+//It is not possible to find a joint where there are only two unknowns. Hence, consider section (1)–(1). 
+//For left hand side part of the frame
+//moment at C
+
+FAE=(RA*CH-PA*CH-PB*CH/2)/(CH)
+printf("\n FAE= %0.0f KN (Tension)",FAE)
+
+//Assuming the directions for FFC and FBC 
+//sum of vertical & sum of horizontal forces is zero
+
+//FFC=FBC*sqrt(2)-RA
+
+FBC=(RA*sin(angleFCH)-PA)/(sqrt(2)*sin(angleFCH)-(1/sqrt(2)))
+printf("\n FBC= %0.2f KN (Comp.)",FBC)
+
+FFC=FBC*sqrt(2)-RA
+printf("\n FFC= %0.2f KN (Tension)",FFC)
+
+//Assumed directions of FBC and FFC are correct. Therefore, FBC is in compression and FFC is in tension. Now we can proceed with method of joints to find the forces in other members. Since it is a symmetric truss, analysis of half the truss is sufficient. Other values may be written down by making use of symmetrry.
+
+//Joint B: sum of forces normal to AC = 0, gives 
+
+FBF=PC*cos(angleACH)
+
+//sum of forces parallel to AC = 0, gives 
+
+FAB=FBC+PC*sin(angleACH)
+
+printf("\n FAB= %0.2f KN (Comp.)",FAB)
+
+
+
+//JOINT A
+//sum of vertical & sum of horizontal forces is zero
+
+FAF=(FAB*sin(angleACH)+PA-RA)/sin(angleFCV)
+
+printf("\n FAF= %0.2f KN (Tension)",FAF)
diff --git a/3862/CH3/EX3.2/Ex3_2.sce b/3862/CH3/EX3.2/Ex3_2.sce
new file mode 100644
index 000000000..91f36f6f2
--- /dev/null
+++ b/3862/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,42 @@
+clear
+//
+
+//Now, we cannot find a joint with only two unknown forces without finding reactions.
+//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero.
+
+//variable declaration
+
+PB=40.0
+PC=50.0
+PE=60.0
+
+theta=60.0
+
+RD=(PC*3+PE*2+PB*1)/(4.0)
+
+RA=PB+PC+PE-RD
+
+FAB=RA/sin(theta*%pi/180)
+
+printf("\n FAB= %0.4f KN (Comp.)",FAB)
+
+FAE=FAB*cos(theta*%pi/180)
+
+printf("\n FAE= %0.4f KN (Tension)",FAE)
+
+FDC=RD/sin(theta*%pi/180)
+
+printf("\n FDC= %0.4f KN (Comp.) ",FDC)
+
+FDE=FDC*cos(theta*%pi/180)
+
+printf("\n FDE= %0.4f KN (Tension) ",FDE)
+
+FBE=(FAB*sin(theta*%pi/180)-PB)/sin(theta*%pi/180)
+
+FBC=(FAB+FBE)*(0.5)
+printf("\n FBC= %0.4f KN (Comp.)",FBC)
+
+
+FCE=(FDC*sin(theta*%pi/180)-PC)/(sin(theta*%pi/180))
+printf("\n FCE= %0.4f KN (Tension)",FCE)
diff --git a/3862/CH3/EX3.3/Ex3_3.sce b/3862/CH3/EX3.3/Ex3_3.sce
new file mode 100644
index 000000000..9c0f84347
--- /dev/null
+++ b/3862/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,53 @@
+clear
+//
+
+//variable declaration
+
+PB=20.0       //Load at point B,KN
+PC=10.0       //Load at point C,KN 
+thetaA=60.0   //angleBAC
+thetaD=30.0   //angleBDC
+
+AC=3.0   //length,m
+CD=3.0   //length,m
+
+AB=(AC+CD)*cos(thetaA*%pi/180)
+BD=(AC+CD)*cos(thetaD*%pi/180)
+//mistake in book
+//angleBCA=angleABC=theta
+
+theta=(180.0-thetaA)/(2.0) 
+
+//Taking moment about A, we get
+RD=(PC*AC+PB*AC*cos(thetaA*%pi/180))/(AC+CD)
+
+RA=PC+PB-RD
+//Joint A
+//vertical & horizontal forces sum to zero
+
+FAB=RA/sin(thetaA*%pi/180)
+
+printf("\n FAB= %0.2f KN  [Comp.]",FAB)
+FAC=FAB*cos(thetaA*%pi/180)
+printf("\n FAC= %0.2f KN  [Tensile]",FAC)
+
+//Joint D
+//vertical & horizontal forces sum to zero
+
+FDB=RD/sin(thetaD*%pi/180)
+
+printf("\n FDB= %0.2f KN  [Comp.]",FDB)
+FDC=FDB*cos(thetaD*%pi/180)
+printf("\n FDC= %0.2f KN [Tensile]",FDC)
+
+//Joint C
+//vertical & horizontal forces sum to zero
+
+FCB=PC/sin(theta*%pi/180)
+
+printf("\n FCB= %0.2f KN  ",FCB)
+
+//CHECK
+
+FCB=(FDC-FAC)/cos(theta*%pi/180)
+printf("\n FCB= %0.2f KN  Checked",FCB)
diff --git a/3862/CH3/EX3.4/Ex3_4.sce b/3862/CH3/EX3.4/Ex3_4.sce
new file mode 100644
index 000000000..6f799932b
--- /dev/null
+++ b/3862/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,68 @@
+clear
+//
+
+//Now, we cannot find a joint with only two unknown forces without finding reactions.
+//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero.
+
+//variable declaration
+
+PB=30.0      //vertical load at point B,KN
+PC=50.0      //vertical load at point C,KN 
+PDv=40.0     //vertical load at point D,KN
+PDh=20.0     //Horizontal load at point D,KN
+PF=30.0      //vertical load at point F,KN
+HA=PDh
+
+RE=(PC*4+PDv*8+PDh*4+PF*4)/(8.0)
+
+VA=PB+PC+PDv+PF-RE
+
+//joint A
+//sum of vertical & sum of horizontal forces is zero.
+
+FAB=VA
+FAF=HA
+
+//joint E
+//sum of vertical & sum of horizontal forces is zero.
+
+FED=RE
+FEF=0
+
+//Joint B: Noting that inclined member is at 45°
+//sum of vertical & sum of horizontal forces is zero.
+
+theta=45.0
+FBF=(VA-PB)/sin(theta*%pi/180)
+
+printf("\n FBF= %0.4f KN (Tension) ",FBF)
+
+FBC=FBF*cos(theta*%pi/180)
+
+printf("\n FBC= %0.4f KN (Comp.) ",FBC)
+
+//Joint C: 
+//sum of vertical & sum of horizontal forces is zero.
+
+
+FCF=PC
+
+printf("\n FCF= %0.4f KN (Comp.) ",FCF)
+
+FCD=FBC
+
+printf("\n FCD= %0.4f KN (Comp.) ",FCD)
+
+//Joint D: Noting that inclined member is at 45°
+//sum of vertical & sum of horizontal forces is zero.
+
+theta=45.0
+FDF=(RE-PDv)/cos(theta*%pi/180)
+
+printf("\n FDF= %0.4f KN (Tensile) ",FDF)
+
+//check
+
+FDF=(FCD+PDh)/cos(theta*%pi/180)
+
+printf("\n FDF= %0.4f KN Checked ",FDF)
diff --git a/3862/CH3/EX3.5/Ex3_5.sce b/3862/CH3/EX3.5/Ex3_5.sce
new file mode 100644
index 000000000..7868fe5bf
--- /dev/null
+++ b/3862/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,74 @@
+clear
+//
+
+//All inclined members have the same inclination to horizontal. Now, length of an inclined member is BF
+
+//variable declaration
+
+PE=20.0
+AF=3.0
+FE=3.0
+AB=4.0
+FD=4.0
+BD=3.0
+CD=4.0
+
+BF=sqrt((AF**2)+(AB**2))
+DE=BF
+BC=DE
+
+//sin(theta)=AB/BF
+//cos(theta)=AF/BF
+
+theta=asin(AB/BF)
+//As soon as a joint is analysed the forces on the joint are marked on members 
+
+//Joint E
+//Consider the equilibrium of the entire frame,Sum of moments about A is zero,Horizontal forces & Vertical forces is zero.
+
+FED=PE/sin(theta)
+printf("\n FED= %0.0f KN (Tension)",FED)
+
+FEF=FED*cos(theta)
+printf("\n FEF= %0.0f KN (Comp.)",FEF)
+
+//At this stage as no other joint is having only two unknowns, no further progress is possible. Let us find the reactions at the supports considering the whole structure. Let the reaction be RC HORIZONTAL at point C,VA,HA at point A Vertically & Horizontally respectively.
+//Taking moment at point A,
+
+RC=PE*6/8 
+//sum of vertical & sun of horizontal forces is zero.
+
+VA=PE
+HA=RC
+
+//Joint A
+//sum of vertical & sun of horizontal forces is zero.
+FAB=VA
+printf("\n FAB= %0.0f KN (Comp.)",FAB)
+
+FAF=HA
+printf("\n FAF= %0.0f KN (Comp.)",FAF)
+
+//Joint C
+//sum of vertical & sun of horizontal forces is zero.
+FCB=RC/cos(theta)
+printf("\n FCB= %0.0f KN (Comp.)",FCB)
+
+FCD=FCB*sin(theta)
+printf("\n FCD= %0.0f KN (Tension)",FCD)
+
+//Joint B
+//sum of vertical & sun of horizontal forces is zero.
+
+FBF=(FCB*sin(theta)-FAB)/sin(theta)
+
+printf("\n FBF= %e ",FBF)
+
+FBD=FCB*cos(theta)
+printf("\n FBD= %0.0f KN (Tension)",FBD)
+
+//joint F
+//sum of vertical & sun of horizontal forces is zero.
+
+FFD=FBF
+printf("\n FFD= %e  ",FFD)
diff --git a/3862/CH3/EX3.6/Ex3_6.sce b/3862/CH3/EX3.6/Ex3_6.sce
new file mode 100644
index 000000000..7f1930dc4
--- /dev/null
+++ b/3862/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,86 @@
+clear
+//
+
+//variable declaration
+
+AB=2.0    //length of beam AB,m
+BD=2.0    //length of beam BD,m
+DF=2.0    //length of beam DF,m
+FH=3.0    //length of beam FH,m
+FG=4.0    //length of beam FG,m
+PF=12.0   //Vertical Load at point F,KN
+PH=20.0   //Vertical Load at point H,KN
+
+//mistake in book FG=4.0 , given FG=2.0 
+
+theta1=atan(FG/(AB+BD+DF))
+theta3=atan(FG/FH)
+theta2=theta3
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+//joint H
+
+FHG=PH/sin(theta3)
+printf("\n FHG= %0.0f KN (Comp.)",FHG)
+
+FHF=FHG*cos(theta2)
+printf("\n FHF= %0.0f KN (Tension)",FHF)
+
+//taking moment at G
+
+RA=PH*FH/(AB+BD+DF)
+
+RG=RA+PF+PH
+
+//joint A
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FAC=RA/sin(theta1)
+printf("\n FAC= %0.4f KN (Comp.)",FAC)
+
+FAB=FAC*cos(theta1)
+printf("\n FAB= %0.0f KN (Tension)",FAB)
+ 
+//joint B
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FBC=0
+printf("\n FBC= %e  ",FBC)
+FBA=FAB
+FBD=FBA
+printf("\n FBD=FBA %0.0f KN (Tension)",FBD)
+ 
+//Joint C: Sum of Forces normal to AC = 0, gives FCD =0 since FBC = 0 ,sum of Forces parallel to CE =0 
+
+FCA=FAC
+FCE=FCA
+printf("\n FCE=FCA %0.4f KN (Comp.)",FCE)
+
+
+//joint D
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FDE=0
+printf("\n FDE= %e  ",FDE)
+
+FDB=FBD
+FDF=FDB
+
+printf("\n FDF=FDB %0.0f KN (Tension)",FDF)
+
+//Joint E: sum of Forces normal to CG = 0, gives FEF = 0 and sum of Forces in the direction of  CG = 0, gives 
+
+FEF=0
+
+
+FEG=FCE
+
+printf("\n FEG=FCE= %0.4f KN (Comp.)",FEG)
+
+//Joint F:
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FFG=PF
+
+printf("\n FFG= %0.0f KN (Tension)",FFG)
diff --git a/3862/CH3/EX3.7/Ex3_7.sce b/3862/CH3/EX3.7/Ex3_7.sce
new file mode 100644
index 000000000..f3534f490
--- /dev/null
+++ b/3862/CH3/EX3.7/Ex3_7.sce
@@ -0,0 +1,99 @@
+clear
+//
+
+// Since all members are 3 m long, all triangles are equilateral and hence all inclined members are at 60° to horizontal. Joint-by-joint analysis is carried out . Then nature of the force is determined. 
+
+//variable declaration
+
+AB=3.0
+BC=AB
+AC=AB
+BD=BC
+CD=BD
+CE=CD
+DE=CE
+EF=DE
+DF=DE
+EG=DE
+FG=DF
+
+theta=60.0*%pi/180 //angles BAC,BCA,DCE,DEC,FEG,FGE,°
+
+PB=40.0    //Vertical Loading at point B,KN
+PD=30.0    //Vertical Loading at point D,KN
+HF=10.0    //Horizontal Loading at point F,KN
+PG=20.0    //Vertical Loading at point G,KN
+
+//joint G
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FGF=PG/sin(theta)
+
+printf("\n FGF= %0.4f KN (Tension)",FGF)
+
+FGE=FGF*cos(theta)
+
+printf("\n FGE= %0.4f KN (Comp.)",FGE)
+
+//joint F
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FFG=FGF
+
+printf("\n FFG= %0.4f KN (Comp.)",FFG)
+
+FFE=FGF
+FFD=FGF*cos(theta)+FFE*cos(theta)-HF
+printf("\n FFD= %0.4f KN (Tension)",FFD)
+
+//Now, without finding reaction we cannot proceed. Hence, consider equilibrium of the entire truss
+//moment about point A
+
+RE=((PB*AC/2)-(HF*EF*sin(theta))+(PD*(AC+CE/2))+(PG*(AC+CE+EG)))/(AC+CE)
+
+VA=PB+PD+PG-RE
+
+HA=HF
+
+//joint A
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FAB=VA/sin(theta)
+
+printf("\n FAB= %0.4f KN (Comp.)",FAB)
+
+FAC=FAB*cos(theta)-HF
+
+printf("\n FAC= %0.4f KN (Tension)",FAC)
+
+
+//joint B
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FBC=(PB-FAB*sin(theta))/sin(theta)
+
+printf("\n FBC= %0.4f KN (Comp.)",FBC)
+
+FBA=FAB
+FBD=-FBC*cos(theta)+FBA*cos(theta)
+
+printf("\n FBD= %0.4f KN (Comp.)",FBD)
+
+//joint C
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FCD=FBC*sin(theta)/sin(theta)
+
+printf("\n FCD= %0.4f KN (Tension)",FCD)
+
+FCE=FCD*cos(theta)+FBC*cos(theta)-FAC
+
+printf("\n FCE= %0.4f KN (Comp.)",FCE)
+
+//joint D
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FDE=(PD+FCD*sin(theta))/sin(theta)
+
+printf("\n FDE= %0.4f KN (Comp.)",FDE)
diff --git a/3862/CH3/EX3.8/Ex3_8.sce b/3862/CH3/EX3.8/Ex3_8.sce
new file mode 100644
index 000000000..acf2e8f75
--- /dev/null
+++ b/3862/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,40 @@
+clear
+//
+
+//Each load is 10 kN and all triangles are equilateral with sides 4 m.
+
+//variable declaration
+
+PB=10.0
+PD=PB
+PF=PD
+AB=4.0
+BC=AB
+AC=BC
+BD=BC
+CD=BC
+DE=CD
+CE=CD
+DF=DE
+EF=DE
+EG=DE
+FG=EF
+//Take section (A)–(A), which cuts the members FH, GH and GI and separates the truss into two parts. 
+AG=AC+CE+EG
+BG=CE+EG+AC/2
+DG=EG+CE/2
+FG1=EG/2
+RA=PB*7/2
+RO=RA
+theta=60.0*%pi/180
+//moment at point G
+FFH=(RA*AG-PB*BG-PD*DG-PF*FG1)/(FG*sin(theta))
+printf("\n FFH= %0.4f KN (Comp.)",FFH)
+
+//sum of all vertical forces & sum of all horizotal forces is zero
+
+FGH=(RA-PB-PD-PF)/(sin(theta))
+printf("\n FGH= %0.4f KN (Comp.)",FGH)
+
+FGI=FFH+FGH*cos(theta)
+printf("\n FGI= %0.4f KN  (Tensile)",FGI)
diff --git a/3862/CH3/EX3.9/Ex3_9.sce b/3862/CH3/EX3.9/Ex3_9.sce
new file mode 100644
index 000000000..f097d8ab8
--- /dev/null
+++ b/3862/CH3/EX3.9/Ex3_9.sce
@@ -0,0 +1,55 @@
+clear
+//
+
+//To determine reactions, consider equilibrium equations
+
+//variable declaration
+//all Vertical loading are in KN
+PL1=200.0               
+PL2=200.0
+PL3=150.0
+PL4=100.0
+PL5=100.0
+
+//length in m
+UL1=6.0
+UL2=8.0
+UL3=9.0
+UL4=UL2
+UL5=UL1
+
+L1=6.0
+L2=6.0
+L3=6.0
+L4=6.0
+L5=6.0
+L6=6.0
+
+//moment at point LO
+
+R2=(PL1*L1+PL2*(L1+L2)+PL3*(L1+L2+L3)+PL4*(L1+L2+L3+L4)+PL5*(L1+L2+L3+L4+L5))/(L1+L2+L3+L4+L5+L6)
+
+R1=PL1+PL2+PL3+PL4+PL5-R2
+
+//Take the section (1)–(1) and consider the right hand side part.
+
+U3U4=sqrt((1**2)+(UL1**2))
+theta1=asin(1/U3U4)
+
+L3U4=sqrt((UL1**2)+(UL2**2))
+theta2=asin(6/L3U4)
+
+//moment at U4
+
+FL3L4=(R2*(L5+L6)-PL4*L4)/UL4
+
+printf("\n FL3L4= %0.1f KN (Tension)",FL3L4)
+
+//moment at L3
+FU4U3=(-PL4*L4-PL5*(L4+L5)+R2*(L4+L5+L6))/(cos(theta1)*UL3)
+printf("\n FU4U3= %0.1f KN (Comp.)",FU4U3)
+
+//sum of horizontal forces 
+FL4L3=FL3L4
+FU4L3=(-FL4L3+FU4U3*cos(theta1))/sin(theta2)
+printf("\n FU4L3= %0.1f KN (Tension)",FU4L3)