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-rwxr-xr-x278/CH4/EX4.1/ex_4_1.sce24
-rwxr-xr-x278/CH4/EX4.10/ex_4_10.sce44
-rwxr-xr-x278/CH4/EX4.11/ex_4_11.sce54
-rwxr-xr-x278/CH4/EX4.12/ex_4_12.sce52
-rwxr-xr-x278/CH4/EX4.13/ex_4_13.sce36
-rwxr-xr-x278/CH4/EX4.14/ex_4_14.sce76
-rwxr-xr-x278/CH4/EX4.15/ex_4_15.sce64
-rwxr-xr-x278/CH4/EX4.16/ex_4_16.sce48
-rwxr-xr-x278/CH4/EX4.17/ex_4_17.sce28
-rwxr-xr-x278/CH4/EX4.18/ex_4_18.sce32
-rwxr-xr-x278/CH4/EX4.19/ex_4_19.sce32
-rwxr-xr-x278/CH4/EX4.2/ex_4_2.sce26
-rwxr-xr-x278/CH4/EX4.3/ex_4_3.sce39
-rwxr-xr-x278/CH4/EX4.4/ex_4_4.sce55
-rwxr-xr-x278/CH4/EX4.5/ex_4_5.sce30
-rwxr-xr-x278/CH4/EX4.6/ex_4_6.sce22
-rwxr-xr-x278/CH4/EX4.7/ex_4_7.sce29
-rwxr-xr-x278/CH4/EX4.8/ex_4_8.sce20
-rwxr-xr-x278/CH4/EX4.9/ex_4_9.sce28
19 files changed, 368 insertions, 371 deletions
diff --git a/278/CH4/EX4.1/ex_4_1.sce b/278/CH4/EX4.1/ex_4_1.sce
index 225a34457..7b2b1a0c5 100755
--- a/278/CH4/EX4.1/ex_4_1.sce
+++ b/278/CH4/EX4.1/ex_4_1.sce
@@ -1,12 +1,12 @@
-//find diameter of the link stock,if permissible tensile stress is not to exceed 75 MPa.
-clc
-//solution
-//given:
-P=50000//N//maximum load is P
-f=75//(N/mm^2)//stress is given
-pi=3.14
-d=sqrt(4*P/(f*pi))//manipulating to get the value of d
-//d=diameter of link stock //using relation f=P/A
-printf("\n\nThe diameter of link stock is,%f mm\n,",d)
-
-
+
+clc
+//solution
+//given:
+P=50000//N//maximum load is P
+f=75//(N/mm^2)//stress is given
+pi=3.14
+d=sqrt(4*P/(f*pi))//manipulating to get the value of d
+//d=diameter of link stock //using relation f=P/A
+printf("\n\nThe diameter of link stock is,%f mm\n,",d)
+
+
diff --git a/278/CH4/EX4.10/ex_4_10.sce b/278/CH4/EX4.10/ex_4_10.sce
index c1c5e8406..4e05ae713 100755
--- a/278/CH4/EX4.10/ex_4_10.sce
+++ b/278/CH4/EX4.10/ex_4_10.sce
@@ -1,22 +1,22 @@
-//calculate 1.)yield stress 2.)ultimate tensile stress 3.) percentage reduction in area 4.) percentage elongation
-clc
-//solution
-//given
-D=12//mm//initial diameter
-l=60//mm//initial length
-L=80//mm//final length
-d=7//mm//final diameter
-Wy=3400//N//yield load
-Wu=6100//N//ultimate load
-pi=3.14
-A=pi*D^2/4//mm^2//initial area of rod
-a=pi*d^2/4//mm^2//final area of rod
-Fy=Wy/A//N/mm^2//yield stress
-Fu=Wu/A//N/mm^2//ultimate stress
-%ria=(A-a)/A*100//percentage reduction in area
-%eil=(L-l)/L*100//percentage elongation in length
-printf("the yield stress is,%f N/mm^2\n",Fy)
-printf("the ultimate stress is ,%f N/mm^2\n",Fu)
-printf("the percentage reduction in area is,%f\n",%ria)
-printf("the percentage increase in length is,%fn",%eil)
-
+
+clc
+//solution
+//given
+D=12//mm//initial diameter
+l=60//mm//initial length
+L=80//mm//final length
+d=7//mm//final diameter
+Wy=3400//N//yield load
+Wu=6100//N//ultimate load
+pi=3.14
+A=pi*D^2/4//mm^2//initial area of rod
+a=pi*d^2/4//mm^2//final area of rod
+Fy=Wy/A//N/mm^2//yield stress
+Fu=Wu/A//N/mm^2//ultimate stress
+%ria=(A-a)/A*100//percentage reduction in area
+%eil=(L-l)/L*100//percentage elongation in length
+printf("the yield stress is,%f N/mm^2\n",Fy)
+printf("the ultimate stress is ,%f N/mm^2\n",Fu)
+printf("the percentage reduction in area is,%f\n",%ria)
+printf("the percentage increase in length is,%fn",%eil)
+
diff --git a/278/CH4/EX4.11/ex_4_11.sce b/278/CH4/EX4.11/ex_4_11.sce
index 689841369..4bbe0ce2c 100755
--- a/278/CH4/EX4.11/ex_4_11.sce
+++ b/278/CH4/EX4.11/ex_4_11.sce
@@ -1,28 +1,28 @@
-//to find increase in length of compound bar and stress produced in steel and copper bar
-clc
-//solution
-//given
-lc=3000//mm//length of steel and copper bar
-lst=3000//mm//length of steel bar
-Ec=105//kN/mm^2//young's modulus of copper
-Est=210//kN/mm^2//young's modulus of steel
-b=25//mm//width
-t=12.5//mm//thickness
-P=50//kN//load applied
-//refer fig 4.14 in book
-//let dl be increace in length of compound bar
-Ast=b*t//mm^2//area of steel bar
-Ac=b*t//mm^2//area of copper bar
-Pc=(P*Ec)/(Ec+Est)//kN//load taken by copper bar
-Pst=P-Pc//kN//load taken by steel bar
-dl=(Pc*lc)/(Ac*Ec)//mm//change in length
-//stresses produced in individual bars are
-//since strain produced are same therefore
-//(Fst/Est)=(Fc/Ec)//since Est=2Ec,therefore Fst(stress in steel)=2*Fc(stress in copper)
-P=Pst+Pc//(Fst*Ast)+(Fc*Ac)//Ast=Ac//Fst=2Fc,therefore gievn equation can ve written as
-//50=2*Fc*Ac+(Fc*Ac)
-Fc=50/(3*Ac)//N/mm^2//stress in copper bar
-Fst=2*Fc//N/mm^2//stress in steel bar
-printf("the change in lentgth of compound bar is,%f mm\n",dl)
-printf("the stress in copper bar is ,%f kN/mm^2\n",Fc)
+
+clc
+//solution
+//given
+lc=3000//mm//length of steel and copper bar
+lst=3000//mm//length of steel bar
+Ec=105//kN/mm^2//young's modulus of copper
+Est=210//kN/mm^2//young's modulus of steel
+b=25//mm//width
+t=12.5//mm//thickness
+P=50//kN//load applied
+//refer fig 4.14 in book
+//let dl be increace in length of compound bar
+Ast=b*t//mm^2//area of steel bar
+Ac=b*t//mm^2//area of copper bar
+Pc=(P*Ec)/(Ec+Est)//kN//load taken by copper bar
+Pst=P-Pc//kN//load taken by steel bar
+dl=(Pc*lc)/(Ac*Ec)//mm//change in length
+//stresses produced in individual bars are
+//since strain produced are same therefore
+//(Fst/Est)=(Fc/Ec)//since Est=2Ec,therefore Fst(stress in steel)=2*Fc(stress in copper)
+P=Pst+Pc//(Fst*Ast)+(Fc*Ac)//Ast=Ac//Fst=2Fc,therefore gievn equation can ve written as
+//50=2*Fc*Ac+(Fc*Ac)
+Fc=50/(3*Ac)//N/mm^2//stress in copper bar
+Fst=2*Fc//N/mm^2//stress in steel bar
+printf("the change in lentgth of compound bar is,%f mm\n",dl)
+printf("the stress in copper bar is ,%f kN/mm^2\n",Fc)
printf("the stress in steel bar is , %f kN/mm^2",Fst) \ No newline at end of file
diff --git a/278/CH4/EX4.12/ex_4_12.sce b/278/CH4/EX4.12/ex_4_12.sce
index 8c7ef529a..042fa0f56 100755
--- a/278/CH4/EX4.12/ex_4_12.sce
+++ b/278/CH4/EX4.12/ex_4_12.sce
@@ -1,27 +1,27 @@
-//find stress after putting the system in lathe
-clc
-//soluton
-//given
-Ds=18//mm//diameter of steel
-Dc1=24//mm//inner diameter of copper rod initially
-Dc2=40//mm//outer diametr of copper
-Fs=10//N/mm^2//stress in steel rod
-pi=3.14
-As=(pi*Ds^2)/4//mm^2//area of steel rod
-Ac=(pi*(Dc2^2-Dc1^2))/4//mm^2//area of copper rod
-//since tensile load on steel rod is equal to compressive load on copper rod,therefore
-//Fs*As=Fc*Ac,therefore
-Fc=Fs*As/Ac//stress in copper rod//N/mm^2
-//when copper rod is reduced outside diametr changes to 40-1.5*2=37mm,therefore new area is
-Ac1=(pi*(37^2-24^2))//mm^2
-//cross section of other half remain same//if Ac2 is the area of remainder then Ac2=Ac
-//let Fc1 be stress in reduced section,Fc2 be stress in remainder ,Fs1 stress in rod aftre turning
-//since load on copper tube is equal to load on steel tube, therefore Ac1*Fc1=Ac2*Fc2=As*Fs1
-//from above equations Fc1=0.41*Fs1,Fc2=0.32*Fs1
-//let L be the length of steela nd copper rod ,since total change in length is equal to change inlength of rduced section before and aftr turning adn change in length of remainder section beofre and aftre turning
-//dl=dl1+dl2
-//(Fs-Fs1)*L/Es=(Fc1-Fc)*L/(2*Ec)+(Fc2-Fc)*(L)/(2*Ec)
-//given Es=2Ec
-//10-Fs1=0.41*Fs1-3.16+0.32*Fs1-3.16
-Fs1=(10+3.16+3.16)/(1+0.41+0.32)
+
+clc
+//soluton
+//given
+Ds=18//mm//diameter of steel
+Dc1=24//mm//inner diameter of copper rod initially
+Dc2=40//mm//outer diametr of copper
+Fs=10//N/mm^2//stress in steel rod
+pi=3.14
+As=(pi*Ds^2)/4//mm^2//area of steel rod
+Ac=(pi*(Dc2^2-Dc1^2))/4//mm^2//area of copper rod
+//since tensile load on steel rod is equal to compressive load on copper rod,therefore
+//Fs*As=Fc*Ac,therefore
+Fc=Fs*As/Ac//stress in copper rod//N/mm^2
+//when copper rod is reduced outside diametr changes to 40-1.5*2=37mm,therefore new area is
+Ac1=(pi*(37^2-24^2))//mm^2
+//cross section of other half remain same//if Ac2 is the area of remainder then Ac2=Ac
+//let Fc1 be stress in reduced section,Fc2 be stress in remainder ,Fs1 stress in rod aftre turning
+//since load on copper tube is equal to load on steel tube, therefore Ac1*Fc1=Ac2*Fc2=As*Fs1
+//from above equations Fc1=0.41*Fs1,Fc2=0.32*Fs1
+//let L be the length of steela nd copper rod ,since total change in length is equal to change inlength of rduced section before and aftr turning adn change in length of remainder section beofre and aftre turning
+//dl=dl1+dl2
+//(Fs-Fs1)*L/Es=(Fc1-Fc)*L/(2*Ec)+(Fc2-Fc)*(L)/(2*Ec)
+//given Es=2Ec
+//10-Fs1=0.41*Fs1-3.16+0.32*Fs1-3.16
+Fs1=(10+3.16+3.16)/(1+0.41+0.32)
printf("the stress in the rod is,%f N/mm^2",Fs1) \ No newline at end of file
diff --git a/278/CH4/EX4.13/ex_4_13.sce b/278/CH4/EX4.13/ex_4_13.sce
index 5fccf8919..b56f36abf 100755
--- a/278/CH4/EX4.13/ex_4_13.sce
+++ b/278/CH4/EX4.13/ex_4_13.sce
@@ -1,19 +1,19 @@
-//find the internal diameter of the tyre and least temp upto which type must be heated
-clc
-//solution
-//given
-D=1200//mm//diameter of wheel
-f=100//N/mm^2//stress
-E=200*10^3//N/mm^2//young's modulus
-a=6.5*10^-6//per degree celcius
-//we know stress/strain=E
-//100/x=E
-x=100/E//
-//x=(D-d)/d
-//x=D/d-1
-d=D/(x+1)//mm
-//let t be least temp to which tyre must be heated
-//pi*D=Pi*d(!+at)
-t=(D-d)/(d*a)
-printf("the internal diameter is, %f mm\n",d)
+
+clc
+//solution
+//given
+D=1200//mm//diameter of wheel
+f=100//N/mm^2//stress
+E=200*10^3//N/mm^2//young's modulus
+a=6.5*10^-6//per degree celcius
+//we know stress/strain=E
+//100/x=E
+x=100/E//
+//x=(D-d)/d
+//x=D/d-1
+d=D/(x+1)//mm
+//let t be least temp to which tyre must be heated
+//pi*D=Pi*d(!+at)
+t=(D-d)/(d*a)
+printf("the internal diameter is, %f mm\n",d)
printf("the least temp is,%f degree celcius",t) \ No newline at end of file
diff --git a/278/CH4/EX4.14/ex_4_14.sce b/278/CH4/EX4.14/ex_4_14.sce
index c6d1bcbbf..a8ab6428d 100755
--- a/278/CH4/EX4.14/ex_4_14.sce
+++ b/278/CH4/EX4.14/ex_4_14.sce
@@ -1,39 +1,39 @@
-//find stress 1.)when supports are unyielding 2.)when supports yield by 0.1 mm
-clc
-//solution
-//given
-t1=37//deg celcius
-t2=20//deg celcius
-Es=210*10^9//N/m^2
-Ed=74*10^9//N/m^2
-as=11.7*10^-6//per degree celcius
-aa=23.4*10^-6//per degree celcius
-ds=0.05//m
-da=0.025//m
-ls=0.6//m
-la=0.3//m
-pi=3.14
-//refer fig4.16 in book
-t=t1-t2//degree celcius
-x1=as*ls*t//contraction in steel bar
-x2=aa*la*t//contaction in aluminium bar
-x=x1+x2//total contraction
-//assume support B is removed,therefore there will no stress in bar,let us assume P force is applied to the right end to brougth in contact with support B..refer ffig 4.17
-As=(pi/4)*ds^2//m^2//area of steel bar
-Aa=(pi/4)*da^2//m^2
-//we know dls=change in length=(P*ls)/(As*Es),therefore dls=P*1.455*10^-9//m
-//dla=P*8.257*10^-9//m
-//threfore total dl=dls + dla=9.712*10^-9 *P//m
-//P*9.712*10^-9=x
-P=x/(9.712*10^-9)
-fs=P/As//stress in steel bar//N/m^2
-fa=P/Aa//stress in aluminiumbar//N/m^2
-//when supports are yielding by 0.1 mm
-X=x-10^-4//m
-P1=X/(9.712*10^-9)//N
-fs1=P1/As//N/m^2
-fa1=P1/Aa//N/m^2
-printf(" the initial stress in steel bar is, %f N/m^2\n",fs)
-printf("the initial stress in aluminium bar is,%f N/m^2\n",fa)
-printf("the final stress in steel bar is,%f N/m^2\n",fs1)
+
+clc
+//solution
+//given
+t1=37//deg celcius
+t2=20//deg celcius
+Es=210*10^9//N/m^2
+Ed=74*10^9//N/m^2
+as=11.7*10^-6//per degree celcius
+aa=23.4*10^-6//per degree celcius
+ds=0.05//m
+da=0.025//m
+ls=0.6//m
+la=0.3//m
+pi=3.14
+//refer fig4.16 in book
+t=t1-t2//degree celcius
+x1=as*ls*t//contraction in steel bar
+x2=aa*la*t//contaction in aluminium bar
+x=x1+x2//total contraction
+//assume support B is removed,therefore there will no stress in bar,let us assume P force is applied to the right end to brougth in contact with support B..refer ffig 4.17
+As=(pi/4)*ds^2//m^2//area of steel bar
+Aa=(pi/4)*da^2//m^2
+//we know dls=change in length=(P*ls)/(As*Es),therefore dls=P*1.455*10^-9//m
+//dla=P*8.257*10^-9//m
+//threfore total dl=dls + dla=9.712*10^-9 *P//m
+//P*9.712*10^-9=x
+P=x/(9.712*10^-9)
+fs=P/As//stress in steel bar//N/m^2
+fa=P/Aa//stress in aluminiumbar//N/m^2
+//when supports are yielding by 0.1 mm
+X=x-10^-4//m
+P1=X/(9.712*10^-9)//N
+fs1=P1/As//N/m^2
+fa1=P1/Aa//N/m^2
+printf(" the initial stress in steel bar is, %f N/m^2\n",fs)
+printf("the initial stress in aluminium bar is,%f N/m^2\n",fa)
+printf("the final stress in steel bar is,%f N/m^2\n",fs1)
printf("the final stress in alu bar is,%f N/m^2",fa1) \ No newline at end of file
diff --git a/278/CH4/EX4.15/ex_4_15.sce b/278/CH4/EX4.15/ex_4_15.sce
index aff2f336c..b7a476b5d 100755
--- a/278/CH4/EX4.15/ex_4_15.sce
+++ b/278/CH4/EX4.15/ex_4_15.sce
@@ -1,33 +1,33 @@
-//find stres induced in both bars and pins
-clc
-//solution
-//given
-dc=0.050//m
-dse=0.075//m
-dsi=0.050//m
-dp=0.018//m
-t=50//degree celcius
-Es=210*10^9//N/m^2
-Ec=105*10^9//N/m^2
-as=11.5*10^-6//per degree celcius
-ac=17*10^-6//per degree celcius
-//refer fig 4.18
-pi=3.14
-Ac=(pi/4)*dc^2//m^2
-As=(pi/4)*(dse^2-dsi^2)//m^2
-Ap=(pi/4)*(dp)^2
-//let l be the length of rods
-//dlc=l*ac*t=850*10^-6*l
-//dls=l*as*t=575*10^-6*l
-//x=dlc-dls=275*10^-6*l
-//x1=(P*l)/(Ac*Ec)=(P*l)/(206.22*10^6)//m
-//x2=(P*l)/(As*Es)=(P*l)/(515.55*10^6)//m
-//therefore X=x1+x2=(6.79*10^-9*P*l)
-//x=X
-P=(275*10^-6)/(6.79*10^-9)//N
-fc=P/Ac//N/m^2
-fs=P/As//N/m^2
-tp=P/(2*Ap)//N/m^2
-printf("the stress in coper bar is, %f N/m^2\n",fc)
-printf("the stress in steel bar is, %f N/m^2\n",fs)
+
+clc
+//solution
+//given
+dc=0.050//m
+dse=0.075//m
+dsi=0.050//m
+dp=0.018//m
+t=50//degree celcius
+Es=210*10^9//N/m^2
+Ec=105*10^9//N/m^2
+as=11.5*10^-6//per degree celcius
+ac=17*10^-6//per degree celcius
+//refer fig 4.18
+pi=3.14
+Ac=(pi/4)*dc^2//m^2
+As=(pi/4)*(dse^2-dsi^2)//m^2
+Ap=(pi/4)*(dp)^2
+//let l be the length of rods
+//dlc=l*ac*t=850*10^-6*l
+//dls=l*as*t=575*10^-6*l
+//x=dlc-dls=275*10^-6*l
+//x1=(P*l)/(Ac*Ec)=(P*l)/(206.22*10^6)//m
+//x2=(P*l)/(As*Es)=(P*l)/(515.55*10^6)//m
+//therefore X=x1+x2=(6.79*10^-9*P*l)
+//x=X
+P=(275*10^-6)/(6.79*10^-9)//N
+fc=P/Ac//N/m^2
+fs=P/As//N/m^2
+tp=P/(2*Ap)//N/m^2
+printf("the stress in coper bar is, %f N/m^2\n",fc)
+printf("the stress in steel bar is, %f N/m^2\n",fs)
printf("the stress is pin is,%f N.m^2",tp) \ No newline at end of file
diff --git a/278/CH4/EX4.16/ex_4_16.sce b/278/CH4/EX4.16/ex_4_16.sce
index 0d9330973..72f17c05f 100755
--- a/278/CH4/EX4.16/ex_4_16.sce
+++ b/278/CH4/EX4.16/ex_4_16.sce
@@ -1,25 +1,25 @@
-//find size of rod
-clc
-//solution
-//given
-P=50*10^3//N//load
-ft=100//N/mm^2//stress
-//let d be diameter in mm
-pi=3.14
-//A=(pi/4)*d^2//area
-//P=ft*A=100*A
-//d^2=(50*1000/78.54)
-d=sqrt(50000/78.54)//mm
-//let x be side if rod is square
-//P=ft*A=100*x^2
-x=sqrt(500)//mm
-//if rod s rectangular
-//A=b*t//b=3t,therfore
-//A=3*t^2//m^2
-//P=ft*A
-t=sqrt(50000/300)//mm
-b=3*t//mm
-printf("the diameter if rod is cylindrical is,%f mm\n",d)
-printf("the side is rod is square is,%f mm\n",x)
-printf("the length if rod is rectangular is ,%f mm \n",t)
+
+clc
+//solution
+//given
+P=50*10^3//N//load
+ft=100//N/mm^2//stress
+//let d be diameter in mm
+pi=3.14
+//A=(pi/4)*d^2//area
+//P=ft*A=100*A
+//d^2=(50*1000/78.54)
+d=sqrt(50000/78.54)//mm
+//let x be side if rod is square
+//P=ft*A=100*x^2
+x=sqrt(500)//mm
+//if rod s rectangular
+//A=b*t//b=3t,therfore
+//A=3*t^2//m^2
+//P=ft*A
+t=sqrt(50000/300)//mm
+b=3*t//mm
+printf("the diameter if rod is cylindrical is,%f mm\n",d)
+printf("the side is rod is square is,%f mm\n",x)
+printf("the length if rod is rectangular is ,%f mm \n",t)
printf("the width if rod is rect is ,%f mm",b) \ No newline at end of file
diff --git a/278/CH4/EX4.17/ex_4_17.sce b/278/CH4/EX4.17/ex_4_17.sce
index f7a55f86a..12cc8fa4f 100755
--- a/278/CH4/EX4.17/ex_4_17.sce
+++ b/278/CH4/EX4.17/ex_4_17.sce
@@ -1,15 +1,15 @@
-//find increase in volume
-clc
-//solution
-//given
-l=2400//mm//length
-A=900//mm^2//area
-P=500000//N//load
-m=1/0.25
-E=0.2*10^6//N/mm^2//young's modulus
-//let dV be change in volume
-V=A*l//mm^3//volume of rod
-st=P/(A*E)//strain
-//dV/V=st*(1-(2/m))
-dV=V*st*(1-(2/m))//mm^3
+
+clc
+//solution
+//given
+l=2400//mm//length
+A=900//mm^2//area
+P=500000//N//load
+m=1/0.25
+E=0.2*10^6//N/mm^2//young's modulus
+//let dV be change in volume
+V=A*l//mm^3//volume of rod
+st=P/(A*E)//strain
+//dV/V=st*(1-(2/m))
+dV=V*st*(1-(2/m))//mm^3
printf("the change in volume is approximately,%f mm^3",dV) \ No newline at end of file
diff --git a/278/CH4/EX4.18/ex_4_18.sce b/278/CH4/EX4.18/ex_4_18.sce
index 2250648fb..9394ffa60 100755
--- a/278/CH4/EX4.18/ex_4_18.sce
+++ b/278/CH4/EX4.18/ex_4_18.sce
@@ -1,17 +1,17 @@
-//find stress corresponding to max inst extesnion and value of unkown weight
-clc
-//solution
-//given
-h=10//mm//height thru which weigth fall
-l=3000//mm//length of bar
-A=600//mm^2//xsection area of bar
-dl=2//mm//change in length of bar
-E=200*10^3//N/mm^2
-//let f be stress
-f=(E*dl)/l//N/mm^2
-//let w be value of unknown weigth
-//we know f=(W/A)*[1+sqrt{1+(2*h*A*E/W/l)}]
-//400/3=(W/600)*[1+sqrt{1+(2*10*600*200*1000/W/3000)}]
-W=6400*100/96//N
-printf("the stress induces is,%f N/mm^2\n",f)
+
+clc
+//solution
+//given
+h=10//mm//height thru which weigth fall
+l=3000//mm//length of bar
+A=600//mm^2//xsection area of bar
+dl=2//mm//change in length of bar
+E=200*10^3//N/mm^2
+//let f be stress
+f=(E*dl)/l//N/mm^2
+//let w be value of unknown weigth
+//we know f=(W/A)*[1+sqrt{1+(2*h*A*E/W/l)}]
+//400/3=(W/600)*[1+sqrt{1+(2*10*600*200*1000/W/3000)}]
+W=6400*100/96//N
+printf("the stress induces is,%f N/mm^2\n",f)
printf("the unknown weigth is,%f N",W) \ No newline at end of file
diff --git a/278/CH4/EX4.19/ex_4_19.sce b/278/CH4/EX4.19/ex_4_19.sce
index 941000f97..152ab9898 100755
--- a/278/CH4/EX4.19/ex_4_19.sce
+++ b/278/CH4/EX4.19/ex_4_19.sce
@@ -1,17 +1,17 @@
-//find the maximum instan stress and elongation
-clc
-//solution
-//given
-d=50//mm//diameter of rod
-l=2500//mm//length of bar
-u=100*10^3//N-mm//shock energy
-E=200*10^3//N/mm^2
-//let f be stress
-pi=3.14
-V=(pi/4)*d^2*l//mm^3
-//u=(f^2*V)/(2*E)
-f=sqrt(u*2*E/V)//N/mm^2
-//let dl be elongation produced
-dl=f*l/E//mm
-printf("the stress produced is,%f N/mm^2\n",f)
+
+clc
+//solution
+//given
+d=50//mm//diameter of rod
+l=2500//mm//length of bar
+u=100*10^3//N-mm//shock energy
+E=200*10^3//N/mm^2
+//let f be stress
+pi=3.14
+V=(pi/4)*d^2*l//mm^3
+//u=(f^2*V)/(2*E)
+f=sqrt(u*2*E/V)//N/mm^2
+//let dl be elongation produced
+dl=f*l/E//mm
+printf("the stress produced is,%f N/mm^2\n",f)
printf("elongation priduces is, %f mm",dl) \ No newline at end of file
diff --git a/278/CH4/EX4.2/ex_4_2.sce b/278/CH4/EX4.2/ex_4_2.sce
index 2b24065d7..b3ef68e04 100755
--- a/278/CH4/EX4.2/ex_4_2.sce
+++ b/278/CH4/EX4.2/ex_4_2.sce
@@ -1,13 +1,13 @@
-//find tensile stress induced in the link at section A-A and B-B,refer fig 4.4
-clc
-//solution
-//given
-P=45000//N//load applied
-A1=45*20//mm^2//area of cross section at link A-A
-//stress in section A-A
-f1=P/A1//(N/mm^2)
-printf("the stress in section A-A is ,%f N/mm^2\n",f1)
-//stress in section B-B
-A2=20*(75-40)//mm^2//area of cross section at link B-B
-f2=P/A2//(N/mm^2)
-printf("the stress in B-B section ,%f N/mm^2",f2)
+
+clc
+//solution
+//given
+P=45000//N//load applied
+A1=45*20//mm^2//area of cross section at link A-A
+//stress in section A-A
+f1=P/A1//(N/mm^2)
+printf("the stress in section A-A is ,%f N/mm^2\n",f1)
+//stress in section B-B
+A2=20*(75-40)//mm^2//area of cross section at link B-B
+f2=P/A2//(N/mm^2)
+printf("the stress in B-B section ,%f N/mm^2",f2)
diff --git a/278/CH4/EX4.3/ex_4_3.sce b/278/CH4/EX4.3/ex_4_3.sce
index 1467c605c..8f502e032 100755
--- a/278/CH4/EX4.3/ex_4_3.sce
+++ b/278/CH4/EX4.3/ex_4_3.sce
@@ -1,21 +1,20 @@
-//to find 1.)diameter of the rods,2.)extension in each rod in the length of 2.5m
-clc
-//solution
-//given
-P=3.5*10^6//N//load applied
-f1=85//(N/mm^2)// safe stress
-E=210*10^3//(N/mm^2)//young's modulus
-l=2.5*10^3//mm
-pi=3.14
-//1)diameter of rod(d)
-//let d be diameter of rods in mm
-//since both rods carries equal load ,therefore load on single rod is
-P1=P/2//N
-d=sqrt(4*P1/(f1*pi))//using f1=P/A//mm
-printf("the diameter of rods is,%f mm\n",d)
-//2)extension in rod
-//let x be extension in rod
-//E=(P1*l)/(A*x)
- //P1/A=f1
-x=(f1*l)/E
+clc
+//solution
+//given
+P=3.5*10^6//N//load applied
+f1=85//(N/mm^2)// safe stress
+E=210*10^3//(N/mm^2)//young's modulus
+l=2.5*10^3//mm
+pi=3.14
+//1)diameter of rod(d)
+//let d be diameter of rods in mm
+//since both rods carries equal load ,therefore load on single rod is
+P1=P/2//N
+d=sqrt(4*P1/(f1*pi))//using f1=P/A//mm
+printf("the diameter of rods is,%f mm\n",d)
+//2)extension in rod
+//let x be extension in rod
+//E=(P1*l)/(A*x)
+ //P1/A=f1
+x=(f1*l)/E
printf("the extension of rod is,%f mm",x) \ No newline at end of file
diff --git a/278/CH4/EX4.4/ex_4_4.sce b/278/CH4/EX4.4/ex_4_4.sce
index 1c07aa2f8..fe871324b 100755
--- a/278/CH4/EX4.4/ex_4_4.sce
+++ b/278/CH4/EX4.4/ex_4_4.sce
@@ -1,29 +1,28 @@
-//1.)calculate the stress on the lower washers before the nuts are tightened,refer fig 4.5
-//2.)what could be the stress in upper and lower washers...
-clc
-//solution
-//given
-d=20//mm
-d1=22//mm
-d2=50//mm
-d3=22//mm
-d4=44//mm
-P1=120000//N
-P2=5000//N
-//1)stress on lower washer before the nuts are tightened
-pi=3.14
-A1=(pi/4)*(d2^2-d1^2)//(mm^2)
-A2=(pi/4)*(d4^2-d3^2)//(mm^2)
-//since load is equally distributed on 4 washers,therfore load Q1=P1/4
-Q1=P1/4//N
-//calculating stress on lower washer
-f2=Q1/A1//(N/mm^2)
-printf("\n the stress on lower washer when nuts are not tightened is,%f N/mm^2\n",f2)
-//2)
-//stres on upper washers
-P2=5000//N
-f3=P2/A2//stress//(N/mm^2)
-printf("the stress on upper washer is,%f N/mm^2\n",f3)
-//stress on lower washer when nuts are tightened
-f4=(Q1+P2)/A1//(N/mm^2)
+
+clc
+//solution
+//given
+d=20//mm
+d1=22//mm
+d2=50//mm
+d3=22//mm
+d4=44//mm
+P1=120000//N
+P2=5000//N
+//1)stress on lower washer before the nuts are tightened
+pi=3.14
+A1=(pi/4)*(d2^2-d1^2)//(mm^2)
+A2=(pi/4)*(d4^2-d3^2)//(mm^2)
+//since load is equally distributed on 4 washers,therfore load Q1=P1/4
+Q1=P1/4//N
+//calculating stress on lower washer
+f2=Q1/A1//(N/mm^2)
+printf("\n the stress on lower washer when nuts are not tightened is,%f N/mm^2\n",f2)
+//2)
+//stres on upper washers
+P2=5000//N
+f3=P2/A2//stress//(N/mm^2)
+printf("the stress on upper washer is,%f N/mm^2\n",f3)
+//stress on lower washer when nuts are tightened
+f4=(Q1+P2)/A1//(N/mm^2)
printf("the stress on lower washer when nuts are tightened is,%f N/mm^2",f4) \ No newline at end of file
diff --git a/278/CH4/EX4.5/ex_4_5.sce b/278/CH4/EX4.5/ex_4_5.sce
index df6edb49a..4fbee1358 100755
--- a/278/CH4/EX4.5/ex_4_5.sce
+++ b/278/CH4/EX4.5/ex_4_5.sce
@@ -1,16 +1,16 @@
-//find the compression of the piston rod
-clc
-//solution
-//given
-d=50//mm//diameter of rod
-l=600//mm//length of rod
-D=400//mm//diameter of piston
-p=0.9//(N/mm^2)//maximum steam pressure
-E=210*10^3//(N/mm^2)//young's modulus
-pi=3.14
-A=(pi/4)*D^2//(mm^2)//area of cross section of piston
-P=A*p//N//max load acting on piston
-a=(pi/4)*d^2//(mm^2)//area of cross section of piston rod
-//let x be the compression
-x=(P*l)/(E*a)//mm
+
+clc
+//solution
+//given
+d=50//mm//diameter of rod
+l=600//mm//length of rod
+D=400//mm//diameter of piston
+p=0.9//(N/mm^2)//maximum steam pressure
+E=210*10^3//(N/mm^2)//young's modulus
+pi=3.14
+A=(pi/4)*D^2//(mm^2)//area of cross section of piston
+P=A*p//N//max load acting on piston
+a=(pi/4)*d^2//(mm^2)//area of cross section of piston rod
+//let x be the compression
+x=(P*l)/(E*a)//mm
printf("the compression in piston rod is,%f mm",x) \ No newline at end of file
diff --git a/278/CH4/EX4.6/ex_4_6.sce b/278/CH4/EX4.6/ex_4_6.sce
index 3784bdb6e..2252c573f 100755
--- a/278/CH4/EX4.6/ex_4_6.sce
+++ b/278/CH4/EX4.6/ex_4_6.sce
@@ -1,12 +1,12 @@
-//calculate the force required to produce cicular black of diameter 60mm in a plate of 5mm thickness
-clc
-//solution
-//given
-d=60//mm
-t=5//mm
-u=350//(N/mm^2)//ultimate stress
-pi=3.14
-A=pi*d*t//(mm^2)//area under shear
-//force required to punch a hole
-P=A*u
+
+clc
+//solution
+//given
+d=60//mm
+t=5//mm
+u=350//(N/mm^2)//ultimate stress
+pi=3.14
+A=pi*d*t//(mm^2)//area under shear
+//force required to punch a hole
+P=A*u
printf("the force required is,%f N",P) \ No newline at end of file
diff --git a/278/CH4/EX4.7/ex_4_7.sce b/278/CH4/EX4.7/ex_4_7.sce
index eae6c4341..7e40bdc76 100755
--- a/278/CH4/EX4.7/ex_4_7.sce
+++ b/278/CH4/EX4.7/ex_4_7.sce
@@ -1,16 +1,15 @@
-//find diameter of bars and of the pin,refer fig 4.8
-clc
-//solution
-//given
-P=80000//N//tensile force applied
-f1=100//(N/mm^2)
-f2=80//(N/mm^2)
-//diameter of bars in mm
-//A1=(pi/4)*D1^2//Area of bar
-//P=f1*(pi/4)*D1^2
-D1=sqrt((4*P)/(f1*pi))
-printf("\nthe diameter of bars is,%f mm\n",D1)
-//diameter of pin
-//A2=(2*pi/4)*D2^2//area of pin
-D2=sqrt((4*P)/(2*pi*f2))
+clc
+//solution
+//given
+P=80000//N//tensile force applied
+f1=100//(N/mm^2)
+f2=80//(N/mm^2)
+//diameter of bars in mm
+//A1=(%pi/4)*D1^2//Area of bar
+//P=f1*(%pi/4)*D1^2
+D1=sqrt((4*P)/(f1*pi))
+printf("\nthe diameter of bars is,%f mm\n",D1)
+//diameter of pin
+//A2=(2*%pi/4)*D2^2//area of pin
+D2=sqrt((4*P)/(2*%pi*f2))
printf("\n the diameter of pin is,%f mm \n",D2) \ No newline at end of file
diff --git a/278/CH4/EX4.8/ex_4_8.sce b/278/CH4/EX4.8/ex_4_8.sce
index 32f49a470..9e1e47006 100755
--- a/278/CH4/EX4.8/ex_4_8.sce
+++ b/278/CH4/EX4.8/ex_4_8.sce
@@ -1,11 +1,11 @@
-//find the crushing stress induced between plates and rivet
-clc
-//solution
-//given
-t=16//thickness//mm
-P=48*10^3//N
-n=2//two plates are given
-d=25//mm
-//stress acting
-f=(P/(d*t*n))//(N/mm^2)
+
+clc
+//solution
+//given
+t=16//thickness//mm
+P=48*10^3//N
+n=2//two plates are given
+d=25//mm
+//stress acting
+f=(P/(d*t*n))//(N/mm^2)
printf("the stress acting is,%f N/mm^2",f) \ No newline at end of file
diff --git a/278/CH4/EX4.9/ex_4_9.sce b/278/CH4/EX4.9/ex_4_9.sce
index 58559fed9..4f79dd5cd 100755
--- a/278/CH4/EX4.9/ex_4_9.sce
+++ b/278/CH4/EX4.9/ex_4_9.sce
@@ -1,14 +1,14 @@
-//find the length of the sliding bearing
-clc
-//solution
-//given
-d=25//mm/diameter
-P=2500//N/force
-p=5//(N/mm^2)//bearing pressure
-//A=l*d =l*25//(mm^2)//projected area of bearing
-//p=P/A//pressure=force/area
-//therefore
-l=(P/(25*5))//mm//length
-printf("the required length is,%f mm",l)
-
-
+
+clc
+//solution
+//given
+d=25//mm/diameter
+P=2500//N/force
+p=5//(N/mm^2)//bearing pressure
+//A=l*d =l*25//(mm^2)//projected area of bearing
+//p=P/A//pressure=force/area
+//therefore
+l=(P/(25*5))//mm//length
+printf("the required length is,%f mm",l)
+
+
generated by cgit (git 2.34.1) at 2025-03-12 20:29:19 +0000