19 files changed, 683 insertions, 691 deletions

diff --git a/278/CH9/EX24.5/ex_24_5.sce b/278/CH9/EX24.5/ex_24_5.sce
index 1ca9da8cb..2b6a4dc63 100755
--- a/278/CH9/EX24.5/ex_24_5.sce
+++ b/278/CH9/EX24.5/ex_24_5.sce
@@ -1,26 +1,26 @@
-//find..

-clc

-//solution

-//given

-P=15000//W

-N=900//rpm

-n=4

-R=0.15//m

-u=0.25

-//let m be the mass

-w=2*%pi*N/60//rad/s

-w1=(3/4)*w//rad/s

-r=0.12//m

-//Pc=m*w^2*r=1066*m//N

-//Ps=m*w1^2*r=600m//N

-T=P*60/(2*%pi*N)//N-m

-//T=u*(Pc-Ps)*R*n=70m

-m=T/70//kg

-printf("mass of shoes is,%f kg\n",m)

-a=%pi/3

-l=R*a*1000//mm

-//A=l*n=157*b//mm^2

-//F=A*p=15.7*b//N

-//15.7*b=Pc-Ps=466m

-b=466*m/(15.7)//mm

+
+clc
+//solution
+//given
+P=15000//W
+N=900//rpm
+n=4
+R=0.15//m
+u=0.25
+//let m be the mass
+w=2*%pi*N/60//rad/s
+w1=(3/4)*w//rad/s
+r=0.12//m
+//Pc=m*w^2*r=1066*m//N
+//Ps=m*w1^2*r=600m//N
+T=P*60/(2*%pi*N)//N-m
+//T=u*(Pc-Ps)*R*n=70m
+m=T/70//kg
+printf("mass of shoes is,%f kg\n",m)
+a=%pi/3
+l=R*a*1000//mm
+//A=l*n=157*b//mm^2
+//F=A*p=15.7*b//N
+//15.7*b=Pc-Ps=466m
+b=466*m/(15.7)//mm
 printf("face width is,%f mm\n",b)
\ No newline at end of file
diff --git a/278/CH9/EX9.1/ex_9_1.sce b/278/CH9/EX9.1/ex_9_1.sce
index 0f985f2f3..d36deaa2d 100755
--- a/278/CH9/EX9.1/ex_9_1.sce
+++ b/278/CH9/EX9.1/ex_9_1.sce
@@ -1,28 +1,28 @@
-//find minimum force per pitch and find actual stresses developed

-clc

-//solution

-//given

-t=15//mm

-d=25//mm

-p=75//mm

-ftu=400//N/mm^2

-tu=320//N/mm^2

-fcu=640//N/mm^2

-pi=3.14

-n=2

-FS=4//factor of safety

-//min foce per pitch which will rupture the joint

-Ptu=(p-d)*t*ftu//N//ultimate teraing reisistance

-Psu=n*(pi/4)*d^2^tu//N//ultimate shear stress

-Pcu=n*d*t*fcu//N//ultimate crushing stress

-//actual stress produced in plates and rivets

-Ac=Ptu/4//N

-//we know

-//Ac=(p-d)*t*fta

-fta=Ac/((p-d)*t)//N/mm^2

-Ta=Ac*4/(n*pi*d^2)//N/mm^2

-fca=Ac/(n*d*t)//N/mm^2

-printf("the min force required is,%f N\n",Ptu)

-printf("the actual tearing stress acting is,%f N/mm^2\n",fta)

-printf("the actual shering stress acting is,%f N/mm^2\n",Ta)

+
+clc
+//solution
+//given
+t=15//mm
+d=25//mm
+p=75//mm
+ftu=400//N/mm^2
+tu=320//N/mm^2
+fcu=640//N/mm^2
+pi=3.14
+n=2
+FS=4//factor of safety
+//min foce per pitch which will rupture the joint
+Ptu=(p-d)*t*ftu//N//ultimate teraing reisistance
+Psu=n*(pi/4)*d^2^tu//N//ultimate shear stress
+Pcu=n*d*t*fcu//N//ultimate crushing stress
+//actual stress produced in plates and rivets
+Ac=Ptu/4//N
+//we know
+//Ac=(p-d)*t*fta
+fta=Ac/((p-d)*t)//N/mm^2
+Ta=Ac*4/(n*pi*d^2)//N/mm^2
+fca=Ac/(n*d*t)//N/mm^2
+printf("the min force required is,%f N\n",Ptu)
+printf("the actual tearing stress acting is,%f N/mm^2\n",fta)
+printf("the actual shering stress acting is,%f N/mm^2\n",Ta)
 printf("thr crushing resistance stress is ,%f N/mm^2",fca)
\ No newline at end of file
diff --git a/278/CH9/EX9.10/ex_9_10.sce b/278/CH9/EX9.10/ex_9_10.sce
index 868ca8ae3..71b9f5d7c 100755
--- a/278/CH9/EX9.10/ex_9_10.sce
+++ b/278/CH9/EX9.10/ex_9_10.sce
@@ -1,64 +1,64 @@
-//give design calculation sfor longitudinal and circumferncial joints

-clc

-//soltuion

-//given

-P=2.5//N/mm^2

-D=1600//mm

-ft=75//N/mm^2

-T=60//N/mm^2

-fc=125//N/mm^2

-//design of longitudinal joint

-t=(P*D)/(2*ft)+1//mm

-d=6*sqrt(t)//m

-pi=3.14

-//choose standard avlue fromtable 9.3

-//let p be pitch

-//Pt=(p-d)*t*ft//N//tearing resistance of plate

-//Pt=(p-34.5)*2100//N

-Ps=4*1.875*(pi/4)*d^2*T+(pi/4)*d^2*T//N//shearing resistance of rivet//N//shearing resistance of rivet

-//Ps=Pt

-//2100*(p-34.5)=Ps

-//p=Ps/(2100)+34.5//mm

-//pmax=C*t+41.28=220//mm

-//since p>pmax,therefore

-p=220//mm

-p'=220/2//mm

-d1=0.2*p + 1.15*d//mm/diatnce between outtr and row and next row

-d2=0.165*p+0.67*d//mm//distance ebtween inner row for zigzag riveting

-t1=0.75*t//mm//thickness of wide strap

-t2=0.625*t//mm//thickness of narrow strap

-m=1.5*d//mm//margin

-Pt=(p-d)*t*ft//N

-Pc=5*d*t*fc//N//crushing resistance of rivet

-P=p*t*ft//N//strength of the unriveted

-//joint may also fail due to combine teARING AND shearing reistance

-Pts=(p-2*d)*t*ft+(pi/4)*d^2*T//N

-//eff=(least of Pt,Ps,Pts)/P

-eff=Pts/P//least is Ps

-//desing for circumferential joint

-//let n be number of rivets

-//shearign resistance of revets=total shearing load acting on circumferential joint

-//n*pi*d^2*T/4=pi*D^2*P/4

-//n=D^2*P/(d^2*T)//89.6 say 90

-n=90

-n1=90/2//number of rivets per row

-//p1=pi(D+t)/n'//

-//p1=113.7,say 140mm standard value'

-p1=140//mm

-effj=(p1-d)/(p1)

-d3=0.33*p1+0.67*d//dis btw rows of rivets for zigzag

-m1=1.5*d

-printf("calcultion for longitudinal joint")

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the thickness of wide strap is,%f mm\n",t1)

-printf("the thickness of narrow strap is,%f mm\n",t2)

-printf("the diameter of rivets is,%f mm",d)

-printf("the margine s,%f mm\n",m)

-printf("the distance btw outer and next row is,%f mm\n",d1)

-printf("the distance btw inner rows is,%f mm\n",d2)

-printf("calculation for circumferencial joint\n")

-printf("the num of rivets is,%f\n",n)

-printf("the number of rivets per rwo for cercumferencial joint is,%f\n",n1)

-printf("the distance btw rows of rivets for zigzag riveting is,%f mm\n",d3)

-printf("the margin is,%f mm",m1)

+
+clc
+//soltuion
+//given
+P=2.5//N/mm^2
+D=1600//mm
+ft=75//N/mm^2
+T=60//N/mm^2
+fc=125//N/mm^2
+//design of longitudinal joint
+t=(P*D)/(2*ft)+1//mm
+d=6*sqrt(t)//m
+pi=3.14
+//choose standard avlue fromtable 9.3
+//let p be pitch
+//Pt=(p-d)*t*ft//N//tearing resistance of plate
+//Pt=(p-34.5)*2100//N
+Ps=4*1.875*(pi/4)*d^2*T+(pi/4)*d^2*T//N//shearing resistance of rivet//N//shearing resistance of rivet
+//Ps=Pt
+//2100*(p-34.5)=Ps
+//p=Ps/(2100)+34.5//mm
+//pmax=C*t+41.28=220//mm
+//since p>pmax,therefore
+p=220//mm
+p'=220/2//mm
+d1=0.2*p + 1.15*d//mm/diatnce between outtr and row and next row
+d2=0.165*p+0.67*d//mm//distance ebtween inner row for zigzag riveting
+t1=0.75*t//mm//thickness of wide strap
+t2=0.625*t//mm//thickness of narrow strap
+m=1.5*d//mm//margin
+Pt=(p-d)*t*ft//N
+Pc=5*d*t*fc//N//crushing resistance of rivet
+P=p*t*ft//N//strength of the unriveted
+//joint may also fail due to combine teARING AND shearing reistance
+Pts=(p-2*d)*t*ft+(pi/4)*d^2*T//N
+//eff=(least of Pt,Ps,Pts)/P
+eff=Pts/P//least is Ps
+//desing for circumferential joint
+//let n be number of rivets
+//shearign resistance of revets=total shearing load acting on circumferential joint
+//n*pi*d^2*T/4=pi*D^2*P/4
+//n=D^2*P/(d^2*T)//89.6 say 90
+n=90
+n1=90/2//number of rivets per row
+//p1=pi(D+t)/n'//
+//p1=113.7,say 140mm standard value'
+p1=140//mm
+effj=(p1-d)/(p1)
+d3=0.33*p1+0.67*d//dis btw rows of rivets for zigzag
+m1=1.5*d
+printf("calcultion for longitudinal joint")
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the thickness of wide strap is,%f mm\n",t1)
+printf("the thickness of narrow strap is,%f mm\n",t2)
+printf("the diameter of rivets is,%f mm",d)
+printf("the margine s,%f mm\n",m)
+printf("the distance btw outer and next row is,%f mm\n",d1)
+printf("the distance btw inner rows is,%f mm\n",d2)
+printf("calculation for circumferencial joint\n")
+printf("the num of rivets is,%f\n",n)
+printf("the number of rivets per rwo for cercumferencial joint is,%f\n",n1)
+printf("the distance btw rows of rivets for zigzag riveting is,%f mm\n",d3)
+printf("the margin is,%f mm",m1)
diff --git a/278/CH9/EX9.11/ex_9_11.sce b/278/CH9/EX9.11/ex_9_11.sce
index 81007bf77..710b2d711 100755
--- a/278/CH9/EX9.11/ex_9_11.sce
+++ b/278/CH9/EX9.11/ex_9_11.sce
@@ -1,38 +1,37 @@
-//design joint

-clc

-//solution

-//given

-b=200//mm

-t=12.5//mm

-ft=80//N/mm^2

-T=65//N/mm^2

-fc=160//N/mm^2

-pi=3.14

-printf("the value of d is,%f mm\n",6*sqrt(t))

-//standard value of d=21.5mm

-d=21.5//mm

-//let n be number of rivets

-Pt=(b-d)*t*ft//N

-Ps=1.75*(pi/4)*d^2*T//N

-Pc=d*t*fc//N

-n=Pt/Ps

-t1=0.75*t//mm

-Pt1=(b-d)*t*ft//N

-Pt2=(b-2*d)*t*ft+Ps//N

-Pt3=(b-2*d)*t*ft+(3*Ps)//N

-Ps5=5*Ps//N//for 5 rivets

-Pc5=5*Pc//N//for 5 rivets

-P=b*t*ft//N

-printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Ps5,Pc5)

-//eff=least(Pt1.Pt2,Pt3,Ps5,Pc5)/P

-eff=Pt1/P//since Pt1 is least

-p=3*d +5//mm//pitch

-m=1.5*d//mm

-d1=2.5*d//mm//dis btw rows of rivets

-printf("the diameter is,%f mm\n",d)

-printf("the nuber of rivets is,%f\n",n)

-printf("the thickness of strap is,%f mm\n",t1)

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the marginl pitch is,%f mm\n",m)

-printf("the dis btw row is,%f mm",d1)

+
+//solution
+//given
+b=200//mm
+t=12.5//mm
+ft=80//N/mm^2
+T=65//N/mm^2
+fc=160//N/mm^2
+pi=3.14
+printf("the value of d is,%f mm\n",6*sqrt(t))
+//standard value of d=21.5mm
+d=21.5//mm
+//let n be number of rivets
+Pt=(b-d)*t*ft//N
+Ps=1.75*(pi/4)*d^2*T//N
+Pc=d*t*fc//N
+n=Pt/Ps
+t1=0.75*t//mm
+Pt1=(b-d)*t*ft//N
+Pt2=(b-2*d)*t*ft+Ps//N
+Pt3=(b-2*d)*t*ft+(3*Ps)//N
+Ps5=5*Ps//N//for 5 rivets
+Pc5=5*Pc//N//for 5 rivets
+P=b*t*ft//N
+printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Ps5,Pc5)
+//eff=least(Pt1.Pt2,Pt3,Ps5,Pc5)/P
+eff=Pt1/P//since Pt1 is least
+p=3*d +5//mm//pitch
+m=1.5*d//mm
+d1=2.5*d//mm//dis btw rows of rivets
+printf("the diameter is,%f mm\n",d)
+printf("the nuber of rivets is,%f\n",n)
+printf("the thickness of strap is,%f mm\n",t1)
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the marginl pitch is,%f mm\n",m)
+printf("the dis btw row is,%f mm",d1)
diff --git a/278/CH9/EX9.12/ex_9_12.sce b/278/CH9/EX9.12/ex_9_12.sce
index 51cf3ca09..280c06975 100755
--- a/278/CH9/EX9.12/ex_9_12.sce
+++ b/278/CH9/EX9.12/ex_9_12.sce
@@ -1,38 +1,38 @@
-//desing an economical joint

-clc

-//solution

-//given

-b=350//mm

-t=20//mm

-ft=90//N/mm^2

-T=60//N/mm^2

-fc=150//N/mm^2

-printf("the value of d is,%f mm\n",6*sqrt(t))

-//d=26.8//mm

-//standard value is d=29mm using table 9.7

-d=29//mm

-Pt=(b-d)*t*ft//N

-Ps=1.75*(pi/4)*d^2*T//N

-Pc=d*t*fc//N

-n=Pt/Ps

-t1=0.75*t//mm

-Pt1=(b-d)*t*ft//N

-Pt2=(b-2*d)*t*ft+Ps//N

-Pt3=(b-3*d)*t*ft+(3*Ps)//N

-Pt4=(b-3*d)*t*ft+(6*Ps)//N

-Ps9=9*Ps//N//for 9 rivets

-Pc9=9*Pc//N//for 9 rivets

-P=b*t*ft//N

-printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Pt4,Ps9,Pc9)

-//eff=least(Pt1.Pt2,Pt3,Pt4,Ps9,Pc9)/P

-eff=Pt1/P//since Pt1 is least

-p=3*d +5//mm//pitch

-m=1.5*d//mm

-d1=2.5*d//mm//dis btw rows of rivets

-printf("the diameter is,%f mm\n",d)

-printf("the nuber of rivets is,%f\n",n)

-printf("the thickness of strap is,%f mm\n",t1)

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the marginl pitch is,%f mm\n",m)

+clc
+//solution
+//given
+b=350//mm
+t=20//mm
+pi = %pi;
+ft=90//N/mm^2
+T=60//N/mm^2
+fc=150//N/mm^2
+printf("the value of d is,%f mm\n",6*sqrt(t))
+//d=26.8//mm
+//standard value is d=29mm using table 9.7
+d=29//mm
+Pt=(b-d)*t*ft//N
+Ps=1.75*(pi/4)*d^2*T//N
+Pc=d*t*fc//N
+n=Pt/Ps
+t1=0.75*t//mm
+Pt1=(b-d)*t*ft//N
+Pt2=(b-2*d)*t*ft+Ps//N
+Pt3=(b-3*d)*t*ft+(3*Ps)//N
+Pt4=(b-3*d)*t*ft+(6*Ps)//N
+Ps9=9*Ps//N//for 9 rivets
+Pc9=9*Pc//N//for 9 rivets
+P=b*t*ft//N
+printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Pt4,Ps9,Pc9)
+//eff=least(Pt1.Pt2,Pt3,Pt4,Ps9,Pc9)/P
+eff=Pt1/P//since Pt1 is least
+p=3*d +5//mm//pitch
+m=1.5*d//mm
+d1=2.5*d//mm//dis btw rows of rivets
+printf("the diameter is,%f mm\n",d)
+printf("the nuber of rivets is,%f\n",n)
+printf("the thickness of strap is,%f mm\n",t1)
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the marginl pitch is,%f mm\n",m)
 printf("the dis btw row is,%f mm",d1)
\ No newline at end of file
diff --git a/278/CH9/EX9.13/ex_9_12.sce b/278/CH9/EX9.13/ex_9_12.sce
index 51cf3ca09..6ae7dccb6 100755
--- a/278/CH9/EX9.13/ex_9_12.sce
+++ b/278/CH9/EX9.13/ex_9_12.sce
@@ -1,38 +1,38 @@
-//desing an economical joint

-clc

-//solution

-//given

-b=350//mm

-t=20//mm

-ft=90//N/mm^2

-T=60//N/mm^2

-fc=150//N/mm^2

-printf("the value of d is,%f mm\n",6*sqrt(t))

-//d=26.8//mm

-//standard value is d=29mm using table 9.7

-d=29//mm

-Pt=(b-d)*t*ft//N

-Ps=1.75*(pi/4)*d^2*T//N

-Pc=d*t*fc//N

-n=Pt/Ps

-t1=0.75*t//mm

-Pt1=(b-d)*t*ft//N

-Pt2=(b-2*d)*t*ft+Ps//N

-Pt3=(b-3*d)*t*ft+(3*Ps)//N

-Pt4=(b-3*d)*t*ft+(6*Ps)//N

-Ps9=9*Ps//N//for 9 rivets

-Pc9=9*Pc//N//for 9 rivets

-P=b*t*ft//N

-printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Pt4,Ps9,Pc9)

-//eff=least(Pt1.Pt2,Pt3,Pt4,Ps9,Pc9)/P

-eff=Pt1/P//since Pt1 is least

-p=3*d +5//mm//pitch

-m=1.5*d//mm

-d1=2.5*d//mm//dis btw rows of rivets

-printf("the diameter is,%f mm\n",d)

-printf("the nuber of rivets is,%f\n",n)

-printf("the thickness of strap is,%f mm\n",t1)

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the marginl pitch is,%f mm\n",m)

+
+clc
+//solution
+//given
+b=350//mm
+t=20//mm
+ft=90//N/mm^2
+T=60//N/mm^2
+fc=150//N/mm^2
+printf("the value of d is,%f mm\n",6*sqrt(t))
+//d=26.8//mm
+//standard value is d=29mm using table 9.7
+d=29//mm
+Pt=(b-d)*t*ft//N
+Ps=1.75*(pi/4)*d^2*T//N
+Pc=d*t*fc//N
+n=Pt/Ps
+t1=0.75*t//mm
+Pt1=(b-d)*t*ft//N
+Pt2=(b-2*d)*t*ft+Ps//N
+Pt3=(b-3*d)*t*ft+(3*Ps)//N
+Pt4=(b-3*d)*t*ft+(6*Ps)//N
+Ps9=9*Ps//N//for 9 rivets
+Pc9=9*Pc//N//for 9 rivets
+P=b*t*ft//N
+printf("the value of forces is,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n,%f N\n",Pt1,Pt2,Pt3,Pt4,Ps9,Pc9)
+//eff=least(Pt1.Pt2,Pt3,Pt4,Ps9,Pc9)/P
+eff=Pt1/P//since Pt1 is least
+p=3*d +5//mm//pitch
+m=1.5*d//mm
+d1=2.5*d//mm//dis btw rows of rivets
+printf("the diameter is,%f mm\n",d)
+printf("the nuber of rivets is,%f\n",n)
+printf("the thickness of strap is,%f mm\n",t1)
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the marginl pitch is,%f mm\n",m)
 printf("the dis btw row is,%f mm",d1)
\ No newline at end of file
diff --git a/278/CH9/EX9.14/ex_9_14.sce b/278/CH9/EX9.14/ex_9_14.sce
index 415d017ec..532512913 100755
--- a/278/CH9/EX9.14/ex_9_14.sce
+++ b/278/CH9/EX9.14/ex_9_14.sce
@@ -1,61 +1,61 @@
-//detremine the size of rivets to be used for joint

-clc

-//refer fig 9.24,9.25

-//solution

-//given

-t=25//mm

-P=50000//N

-e=400//mm

-n=7

-T=65//N/mm^2

-fc=120//N/mm^2

-//let xb and yb be center of gravity

-//xb=(x1+x2+x3+x4+x5+x6+x7)/7

-xb=(100+200+200+200)/7//mm

-//yb=(y1+y2+y3+y4+y5+y6+y7)/7

-yb=(200+200+200+100+100)/7//mm

-Ps=P/n

-T1=P*e//turning moment due to P//N-mm

-//l1=l3

-l1=sqrt(100^2+(200-yb)^2)//mm

-l3=sqrt(100^2+(200-yb)^2)//mm

-l2=200-yb//mm

-//l4=l7

-l4=sqrt(100^2+(yb-100)^2)//mm

-l7=sqrt(100^2+(yb-100)^2)//mm

-//l5=l6

-l5=sqrt(100^2+yb^2)//mm

-l6=sqrt(100^2+yb^2)//mm

-//eqauting the moments equal to each other 

-//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2]

-F1=(P*e*l1)/(l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2)//N

-F2=F1*l2/l1//N

-F3=F1*l3/l1//N

-F4=F1*l4/l1//N

-F5=F1*l5/l1//N

-F6=F1*l6/l1//N

-F7=F1*l7/l1//N

-//cos(q1)=100/l3=0.76=a

-//cos(q4)=100/l4=0.99=b

-//cos(q5)=100/l5=0.658=c

-a=0.76

-b=0.99

-c=0.658

-R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)

-R4=sqrt(Ps^2+F4^2+2*F4*Ps*b)

-R5=sqrt(Ps^2+F5^2+2*F5*Ps*c)

-printf("the value R3,R4,R5 are respctively,%f N\n,%f N\n,%f N\n",R3,R4,R5)

-//let d be diameter 

-pi=3.14

-//from above we see that max lod is R5,therefore R5=P

-//R5=(pi/4)*d^2*T

-d=sqrt(R5*4/(pi*T))//mm

-Lc=R5/(d*t)//max crushing load

-printf("the  cordinates of centre of gravity are,%f mm\n,%f mm \n",xb,yb)

-printf("the direct load is,%f N\n",Ps)

-printf("the turning moment is,%f N-mm\n",T1)

-printf("the values of Li respectively is,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",l1,l2,l3,l4,l5,l6,l7)

-printf("the shear loads(Forces F) acting are,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",F1,F2,F3,F4,F5,F6,F7)

-printf("the crushing stress is,%f N/mm^2\n ",Lc)

-printf("the diameter is ,%f mm\n",d)

-printf("since crushing load calculted is less then 120 N/mm^2,therefore desing is safe ")

+
+clc
+//refer fig 9.24,9.25
+//solution
+//given
+t=25//mm
+P=50000//N
+e=400//mm
+n=7
+T=65//N/mm^2
+fc=120//N/mm^2
+//let xb and yb be center of gravity
+//xb=(x1+x2+x3+x4+x5+x6+x7)/7
+xb=(100+200+200+200)/7//mm
+//yb=(y1+y2+y3+y4+y5+y6+y7)/7
+yb=(200+200+200+100+100)/7//mm
+Ps=P/n
+T1=P*e//turning moment due to P//N-mm
+//l1=l3
+l1=sqrt(100^2+(200-yb)^2)//mm
+l3=sqrt(100^2+(200-yb)^2)//mm
+l2=200-yb//mm
+//l4=l7
+l4=sqrt(100^2+(yb-100)^2)//mm
+l7=sqrt(100^2+(yb-100)^2)//mm
+//l5=l6
+l5=sqrt(100^2+yb^2)//mm
+l6=sqrt(100^2+yb^2)//mm
+//eqauting the moments equal to each other 
+//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2]
+F1=(P*e*l1)/(l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2)//N
+F2=F1*l2/l1//N
+F3=F1*l3/l1//N
+F4=F1*l4/l1//N
+F5=F1*l5/l1//N
+F6=F1*l6/l1//N
+F7=F1*l7/l1//N
+//cos(q1)=100/l3=0.76=a
+//cos(q4)=100/l4=0.99=b
+//cos(q5)=100/l5=0.658=c
+a=0.76
+b=0.99
+c=0.658
+R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)
+R4=sqrt(Ps^2+F4^2+2*F4*Ps*b)
+R5=sqrt(Ps^2+F5^2+2*F5*Ps*c)
+printf("the value R3,R4,R5 are respctively,%f N\n,%f N\n,%f N\n",R3,R4,R5)
+//let d be diameter 
+pi=3.14
+//from above we see that max lod is R5,therefore R5=P
+//R5=(pi/4)*d^2*T
+d=sqrt(R5*4/(pi*T))//mm
+Lc=R5/(d*t)//max crushing load
+printf("the  cordinates of centre of gravity are,%f mm\n,%f mm \n",xb,yb)
+printf("the direct load is,%f N\n",Ps)
+printf("the turning moment is,%f N-mm\n",T1)
+printf("the values of Li respectively is,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",l1,l2,l3,l4,l5,l6,l7)
+printf("the shear loads(Forces F) acting are,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",F1,F2,F3,F4,F5,F6,F7)
+printf("the crushing stress is,%f N/mm^2\n ",Lc)
+printf("the diameter is ,%f mm\n",d)
+printf("since crushing load calculted is less then 120 N/mm^2,therefore desing is safe ")
diff --git a/278/CH9/EX9.15/ex_9_14.sce b/278/CH9/EX9.15/ex_9_14.sce
index 415d017ec..532512913 100755
--- a/278/CH9/EX9.15/ex_9_14.sce
+++ b/278/CH9/EX9.15/ex_9_14.sce
@@ -1,61 +1,61 @@
-//detremine the size of rivets to be used for joint

-clc

-//refer fig 9.24,9.25

-//solution

-//given

-t=25//mm

-P=50000//N

-e=400//mm

-n=7

-T=65//N/mm^2

-fc=120//N/mm^2

-//let xb and yb be center of gravity

-//xb=(x1+x2+x3+x4+x5+x6+x7)/7

-xb=(100+200+200+200)/7//mm

-//yb=(y1+y2+y3+y4+y5+y6+y7)/7

-yb=(200+200+200+100+100)/7//mm

-Ps=P/n

-T1=P*e//turning moment due to P//N-mm

-//l1=l3

-l1=sqrt(100^2+(200-yb)^2)//mm

-l3=sqrt(100^2+(200-yb)^2)//mm

-l2=200-yb//mm

-//l4=l7

-l4=sqrt(100^2+(yb-100)^2)//mm

-l7=sqrt(100^2+(yb-100)^2)//mm

-//l5=l6

-l5=sqrt(100^2+yb^2)//mm

-l6=sqrt(100^2+yb^2)//mm

-//eqauting the moments equal to each other 

-//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2]

-F1=(P*e*l1)/(l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2)//N

-F2=F1*l2/l1//N

-F3=F1*l3/l1//N

-F4=F1*l4/l1//N

-F5=F1*l5/l1//N

-F6=F1*l6/l1//N

-F7=F1*l7/l1//N

-//cos(q1)=100/l3=0.76=a

-//cos(q4)=100/l4=0.99=b

-//cos(q5)=100/l5=0.658=c

-a=0.76

-b=0.99

-c=0.658

-R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)

-R4=sqrt(Ps^2+F4^2+2*F4*Ps*b)

-R5=sqrt(Ps^2+F5^2+2*F5*Ps*c)

-printf("the value R3,R4,R5 are respctively,%f N\n,%f N\n,%f N\n",R3,R4,R5)

-//let d be diameter 

-pi=3.14

-//from above we see that max lod is R5,therefore R5=P

-//R5=(pi/4)*d^2*T

-d=sqrt(R5*4/(pi*T))//mm

-Lc=R5/(d*t)//max crushing load

-printf("the  cordinates of centre of gravity are,%f mm\n,%f mm \n",xb,yb)

-printf("the direct load is,%f N\n",Ps)

-printf("the turning moment is,%f N-mm\n",T1)

-printf("the values of Li respectively is,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",l1,l2,l3,l4,l5,l6,l7)

-printf("the shear loads(Forces F) acting are,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",F1,F2,F3,F4,F5,F6,F7)

-printf("the crushing stress is,%f N/mm^2\n ",Lc)

-printf("the diameter is ,%f mm\n",d)

-printf("since crushing load calculted is less then 120 N/mm^2,therefore desing is safe ")

+
+clc
+//refer fig 9.24,9.25
+//solution
+//given
+t=25//mm
+P=50000//N
+e=400//mm
+n=7
+T=65//N/mm^2
+fc=120//N/mm^2
+//let xb and yb be center of gravity
+//xb=(x1+x2+x3+x4+x5+x6+x7)/7
+xb=(100+200+200+200)/7//mm
+//yb=(y1+y2+y3+y4+y5+y6+y7)/7
+yb=(200+200+200+100+100)/7//mm
+Ps=P/n
+T1=P*e//turning moment due to P//N-mm
+//l1=l3
+l1=sqrt(100^2+(200-yb)^2)//mm
+l3=sqrt(100^2+(200-yb)^2)//mm
+l2=200-yb//mm
+//l4=l7
+l4=sqrt(100^2+(yb-100)^2)//mm
+l7=sqrt(100^2+(yb-100)^2)//mm
+//l5=l6
+l5=sqrt(100^2+yb^2)//mm
+l6=sqrt(100^2+yb^2)//mm
+//eqauting the moments equal to each other 
+//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2]
+F1=(P*e*l1)/(l1^2+l2^2+l3^2+l4^2+l5^2+l6^2+l7^2)//N
+F2=F1*l2/l1//N
+F3=F1*l3/l1//N
+F4=F1*l4/l1//N
+F5=F1*l5/l1//N
+F6=F1*l6/l1//N
+F7=F1*l7/l1//N
+//cos(q1)=100/l3=0.76=a
+//cos(q4)=100/l4=0.99=b
+//cos(q5)=100/l5=0.658=c
+a=0.76
+b=0.99
+c=0.658
+R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)
+R4=sqrt(Ps^2+F4^2+2*F4*Ps*b)
+R5=sqrt(Ps^2+F5^2+2*F5*Ps*c)
+printf("the value R3,R4,R5 are respctively,%f N\n,%f N\n,%f N\n",R3,R4,R5)
+//let d be diameter 
+pi=3.14
+//from above we see that max lod is R5,therefore R5=P
+//R5=(pi/4)*d^2*T
+d=sqrt(R5*4/(pi*T))//mm
+Lc=R5/(d*t)//max crushing load
+printf("the  cordinates of centre of gravity are,%f mm\n,%f mm \n",xb,yb)
+printf("the direct load is,%f N\n",Ps)
+printf("the turning moment is,%f N-mm\n",T1)
+printf("the values of Li respectively is,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",l1,l2,l3,l4,l5,l6,l7)
+printf("the shear loads(Forces F) acting are,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n,%f mm\n",F1,F2,F3,F4,F5,F6,F7)
+printf("the crushing stress is,%f N/mm^2\n ",Lc)
+printf("the diameter is ,%f mm\n",d)
+printf("since crushing load calculted is less then 120 N/mm^2,therefore desing is safe ")
diff --git a/278/CH9/EX9.16/ex_9_16.sce b/278/CH9/EX9.16/ex_9_16.sce
index afb4a9cb1..7716e15a2 100755
--- a/278/CH9/EX9.16/ex_9_16.sce
+++ b/278/CH9/EX9.16/ex_9_16.sce
@@ -1,32 +1,32 @@
-//find value of P

-clc

-//solution

-//given

-//refer fig 9.29 and 9.30

-T=100//N/mm^2

-n=4

-d=20//mm

-//Ps=P/4=0.25*P//N

-e=100//mm

-//T1=P*e//turning moment

-//la=ld=200=100//mm

-//lb=lc=100//mm

-//eqauting the moments equal to each other 

-//P*e=(Fa/la)*[la^2+lb^2+lc^2+ld^2]

-//P*e=(Fa/la)*[2*300^2+2*100^2]

-//Fa=P*100*3/2000//N

-//Fa=0.15*P//N

-//Fb=Fa*lb/la=0.05*P//N

-//Fc=Fa*lc/la=0.05*P//N

-//Fd=Fa*ld/la=0.15*P//N

-//Ra=Ps-Fa=0.1*P

-//Rb=Ps-Fb=0.20*P

-//Rc=Ps+Fc=0.30*p

-//Rd=Ps+Fd=0.40*P//N

-//max load is Rd

-//therfore

-pi=3.14

-//Rd=(pi/4)*T*d^2

-//0.40*P=(pi/4)*T*d^2

-P=(pi/4)*T*d^2/0.40

+
+clc
+//solution
+//given
+//refer fig 9.29 and 9.30
+T=100//N/mm^2
+n=4
+d=20//mm
+//Ps=P/4=0.25*P//N
+e=100//mm
+//T1=P*e//turning moment
+//la=ld=200=100//mm
+//lb=lc=100//mm
+//eqauting the moments equal to each other 
+//P*e=(Fa/la)*[la^2+lb^2+lc^2+ld^2]
+//P*e=(Fa/la)*[2*300^2+2*100^2]
+//Fa=P*100*3/2000//N
+//Fa=0.15*P//N
+//Fb=Fa*lb/la=0.05*P//N
+//Fc=Fa*lc/la=0.05*P//N
+//Fd=Fa*ld/la=0.15*P//N
+//Ra=Ps-Fa=0.1*P
+//Rb=Ps-Fb=0.20*P
+//Rc=Ps+Fc=0.30*p
+//Rd=Ps+Fd=0.40*P//N
+//max load is Rd
+//therfore
+pi=3.14
+//Rd=(pi/4)*T*d^2
+//0.40*P=(pi/4)*T*d^2
+P=(pi/4)*T*d^2/0.40
 printf("the value of force P is,%f N",P)
\ No newline at end of file
diff --git a/278/CH9/EX9.17/ex_9_17.sce b/278/CH9/EX9.17/ex_9_17.sce
index 193680bdb..3c4d96e69 100755
--- a/278/CH9/EX9.17/ex_9_17.sce
+++ b/278/CH9/EX9.17/ex_9_17.sce
@@ -1,37 +1,37 @@
-//find diameter  of rivet

-clc

-//solution

-//given

-n=6

-P=60000//N

-e=200//mm

-T=150//N/mm^2

-Ps=P/n

-//l1=l3=l4=l6

-l1=sqrt(75^2+50^2)//mm

-l3=sqrt(75^2+50^2)//mm

-l4=sqrt(75^2+50^2)//mm

-l6=sqrt(75^2+50^2)//mm

-l2=50//mm

-l5=50//mm

-//eqauting the moments equal to each other 

-//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2]

-//P*e=(F1/l1)*[4*l1^2+2*l2^2]

-F1=(P*e*l1)/(4*l1^2+2*l2^2)//N

-F2=F1*l2/l1//N

-F3=F1*l3/l1//N

-F4=F1*l4/l1//N

-F5=F1*l5/l1//N

-F6=F1*l6/l1//N

-//cos(q1)=50/l1=0.555=a

-a=0.555

-R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)

-R2=Ps+F2//N

-printf("the value of forces is,%f N\n,%f N\n",R2,R3)

-//R3>R2

-pi=3.14

-P=(pi/4)*d^2*T

-//R3=P

-d=sqrt(R3/117.8)//mm

-printf("the value of diameter is,%f mm\n",d)

-printf("the standard diameter of is 19.5 mm ")

+
+clc
+//solution
+//given
+n=6
+P=60000//N
+e=200//mm
+T=150//N/mm^2
+Ps=P/n
+//l1=l3=l4=l6
+l1=sqrt(75^2+50^2)//mm
+l3=sqrt(75^2+50^2)//mm
+l4=sqrt(75^2+50^2)//mm
+l6=sqrt(75^2+50^2)//mm
+l2=50//mm
+l5=50//mm
+//eqauting the moments equal to each other 
+//P*e=(F1/l1)*[l1^2+l2^2+l3^2+l4^2+l5^2+l6^2]
+//P*e=(F1/l1)*[4*l1^2+2*l2^2]
+F1=(P*e*l1)/(4*l1^2+2*l2^2)//N
+F2=F1*l2/l1//N
+F3=F1*l3/l1//N
+F4=F1*l4/l1//N
+F5=F1*l5/l1//N
+F6=F1*l6/l1//N
+//cos(q1)=50/l1=0.555=a
+a=0.555
+R3=sqrt(Ps^2+F3^2+2*F3*Ps*a)
+R2=Ps+F2//N
+printf("the value of forces is,%f N\n,%f N\n",R2,R3)
+//R3>R2
+pi=3.14
+P=(pi/4)*d^2*T
+//R3=P
+d=sqrt(R3/117.8)//mm
+printf("the value of diameter is,%f mm\n",d)
+printf("the standard diameter of is 19.5 mm ")
diff --git a/278/CH9/EX9.18/ex_9_18.sce b/278/CH9/EX9.18/ex_9_18.sce
index 37c14bdee..c199d5153 100755
--- a/278/CH9/EX9.18/ex_9_18.sce
+++ b/278/CH9/EX9.18/ex_9_18.sce
@@ -1,48 +1,48 @@
-//determine diametr of diameter and thickness of plate

-clc

-//solution

-//given'

-n=4

-Ab=60//mm

-Cd=60//mm

-Bc=60//mm

-P=100000//N

-Ef=150//mm

-q=30//deg

-Ty=240//N/mm^2

-Fs=1.5

-Fb=125//N/mm^2

-b=240//mm

-//let d be diameter of rivet

-Ps=P/n//N

-e=Ef*sin(q)//mm

-la=60+30//mm

-ld=90//mm

-//la=ld

-//lb=lc

-lb=30//mm

-lc=30//mm

-//eqauting the moments equal to each other 

-//P*e=(Fa/la)*[la^2+lb^2+lc^2+ld^2]

-//10000*75=(Fa/la)*[2*90^2+2*30^2]

-Fa=7500*1000*la/(2*90^2+2*30^2)//N

-Fb=Fa*lb/la//N

-Fc=Fa*lc/la//N

-Fd=Fa*ld/la//N

-a=-sqrt(3)/2//deg

-b=-sqrt(3)/2//deg

-c=sqrt(3)/2

-d=sqrt(3)/2

-Ra=sqrt(Ps^2+ Fa^2+ 2*Fa*Ps*a)

-Rb=sqrt(Ps^2+ Fb^2+ 2*Fb*Ps*b)

-Rc=sqrt(Ps^2+ Fc^2+ 2*Fc*Ps*c)

-Rd=sqrt(Ps^2+ Fd^2+ 2*Fd*Ps*d)

-printf("the value of Ps is,%f N\n ",Ps)

-printf("the value fo forces  rae,%f N\n,%f n\n,%f n\n,%f N\n",Fa,Fb,Fc,Fd)

-printf("the value of Ra,Rb,Rc and Rd are,%f N\n,%f N\n,%f N\n,%f N\n",Ra,Rb,Rc,Rd)

-//since greatest is Rd,therfore Rd=P

-pi=3.14

-//P=(pi/4)*d^2*Ty/Fs//N

-d1=sqrt(Rd/125.7)

-printf("the diametr of rivet is %f mm\n",d1)

-printf("choosing th standard value od d as 23.5 mm\n")

+
+clc
+//solution
+//given'
+n=4
+Ab=60//mm
+Cd=60//mm
+Bc=60//mm
+P=100000//N
+Ef=150//mm
+q=30//deg
+Ty=240//N/mm^2
+Fs=1.5
+Fb=125//N/mm^2
+b=240//mm
+//let d be diameter of rivet
+Ps=P/n//N
+e=Ef*sin(q)//mm
+la=60+30//mm
+ld=90//mm
+//la=ld
+//lb=lc
+lb=30//mm
+lc=30//mm
+//eqauting the moments equal to each other 
+//P*e=(Fa/la)*[la^2+lb^2+lc^2+ld^2]
+//10000*75=(Fa/la)*[2*90^2+2*30^2]
+Fa=7500*1000*la/(2*90^2+2*30^2)//N
+Fb=Fa*lb/la//N
+Fc=Fa*lc/la//N
+Fd=Fa*ld/la//N
+a=-sqrt(3)/2//deg
+b=-sqrt(3)/2//deg
+c=sqrt(3)/2
+d=sqrt(3)/2
+Ra=sqrt(Ps^2+ Fa^2+ 2*Fa*Ps*a)
+Rb=sqrt(Ps^2+ Fb^2+ 2*Fb*Ps*b)
+Rc=sqrt(Ps^2+ Fc^2+ 2*Fc*Ps*c)
+Rd=sqrt(Ps^2+ Fd^2+ 2*Fd*Ps*d)
+printf("the value of Ps is,%f N\n ",Ps)
+printf("the value fo forces  rae,%f N\n,%f n\n,%f n\n,%f N\n",Fa,Fb,Fc,Fd)
+printf("the value of Ra,Rb,Rc and Rd are,%f N\n,%f N\n,%f N\n,%f N\n",Ra,Rb,Rc,Rd)
+//since greatest is Rd,therfore Rd=P
+pi=3.14
+//P=(pi/4)*d^2*Ty/Fs//N
+d1=sqrt(Rd/125.7)
+printf("the diametr of rivet is %f mm\n",d1)
+printf("choosing th standard value od d as 23.5 mm\n")
diff --git a/278/CH9/EX9.2/ex_9_2.sce b/278/CH9/EX9.2/ex_9_2.sce
index 8416bfd1c..a7a89799c 100755
--- a/278/CH9/EX9.2/ex_9_2.sce
+++ b/278/CH9/EX9.2/ex_9_2.sce
@@ -1,27 +1,27 @@
-//find the efficiency of following rivet joints

-clc

-//solution

-//given

-t=6//mm

-d=20//mm

-ft=120//N/mm^2

-T=90//N/mm^2

-fc=180//N/mm^2

-p=50//mm

-pi=3.14

-Pt=(p-d)*t*ft//N//tearing resistance of plate

-Ps=(pi/4)*d^2*T//N//shearing resistance of rivet

-Pc=d*t*fc//N//crushing resistance of rivet

-P=p*t*ft//N//strength of the unriveted

-//eff=(least of Pt,Ps,Pc)/P

-eff=Pt/P//least is Pt

-p1=65//mm

-Pt1=(p1-d)*t*ft//N

-Ps1=(2*pi/4)*d^2*T//N

-Pc1=2*d*t*fc//N

-P2=p1*t*ft//N

-printf("the value of forces are,%f N\n,%f N\n,%f N\n",Pt1,Ps1,Pc1)

-//eff1=least of Pt1,Ps1,Pc1/P2

-eff1=Pt1/P2//least is Pt1

-printf("the efficiency is first case is,%f\n",eff)

+
+clc
+//solution
+//given
+t=6//mm
+d=20//mm
+ft=120//N/mm^2
+T=90//N/mm^2
+fc=180//N/mm^2
+p=50//mm
+pi=3.14
+Pt=(p-d)*t*ft//N//tearing resistance of plate
+Ps=(pi/4)*d^2*T//N//shearing resistance of rivet
+Pc=d*t*fc//N//crushing resistance of rivet
+P=p*t*ft//N//strength of the unriveted
+//eff=(least of Pt,Ps,Pc)/P
+eff=Pt/P//least is Pt
+p1=65//mm
+Pt1=(p1-d)*t*ft//N
+Ps1=(2*pi/4)*d^2*T//N
+Pc1=2*d*t*fc//N
+P2=p1*t*ft//N
+printf("the value of forces are,%f N\n,%f N\n,%f N\n",Pt1,Ps1,Pc1)
+//eff1=least of Pt1,Ps1,Pc1/P2
+eff1=Pt1/P2//least is Pt1
+printf("the efficiency is first case is,%f\n",eff)
 printf("the eff is second case is,%f",eff1)
\ No newline at end of file
diff --git a/278/CH9/EX9.3/ex_9_3.sce b/278/CH9/EX9.3/ex_9_3.sce
index 7e5f4a5ef..e4978f85f 100755
--- a/278/CH9/EX9.3/ex_9_3.sce
+++ b/278/CH9/EX9.3/ex_9_3.sce
@@ -1,18 +1,18 @@
-//find efficiency of joint

-clc

-//solution

-//given

-t=10//mm

-d=25//mm

-p=100//mm

-ft=120//N/mm^2

-T=100//N/mm^2

-fc=150//N/mm^2

-pi=3.14

-Pt=(p-d)*t*ft//N//tearing resistance of plate

-Ps=(2*pi/4)*d^2*T//N//shearing resistance of rivet

-Pc=2*d*t*fc//N//crushing resistance of rivet

-P=p*t*ft//N//strength of the unriveted

-//eff=(least of Pt,Ps,Pc)/P

-eff=Pc/P//least is Pc

+
+clc
+//solution
+//given
+t=10//mm
+d=25//mm
+p=100//mm
+ft=120//N/mm^2
+T=100//N/mm^2
+fc=150//N/mm^2
+pi=3.14
+Pt=(p-d)*t*ft//N//tearing resistance of plate
+Ps=(2*pi/4)*d^2*T//N//shearing resistance of rivet
+Pc=2*d*t*fc//N//crushing resistance of rivet
+P=p*t*ft//N//strength of the unriveted
+//eff=(least of Pt,Ps,Pc)/P
+eff=Pc/P//least is Pc
 printf("the eff is,%f",eff)
\ No newline at end of file
diff --git a/278/CH9/EX9.4/ex_9_4.sce b/278/CH9/EX9.4/ex_9_4.sce
index ba8a8e84a..f7a0a38c4 100755
--- a/278/CH9/EX9.4/ex_9_4.sce
+++ b/278/CH9/EX9.4/ex_9_4.sce
@@ -1,28 +1,28 @@
-//find efficiency

-clc

-//solution

-//given

-t=13//mm

-ft=80//N/mm^2

-T=60//N/mm^2

-fc=120//N/mm^2

-pi=3.14

-d=6*sqrt(t)//mm//dia of rivet

-//use standard value from table 9.3

-//let p be the picth of rivets

-//Pt=(p-d)*t*ft=(p-23)*1040//N//tearing resistance of plate

-Ps=2*(pi/4)*d^2*T//N//shearing resistance of rivet

-//p-23=Ps/1040

-p=23+(Ps/1040)//mm

-//check the limits,if p<=pmax..then it is safe design

-//pmax=C*t+41.28//mm=75.28mm which is more then p

-pb=0.33*p+ 0.67*d//distance btw rivets

-m=1.58*d//margin

-Pt=(p-d)*t*ft//N//tearing resistance of plate

-Ps=(2*pi/4)*d^2*T//N//shearing resistance of rivet

-Pc=2*d*t*fc//N//crushing resistance of rivet

-P=p*t*ft//N//strength of the unriveted

-//eff=(least of Pt,Ps,Pc)/P

-printf("the value of forces are,%f N\n,%f N\n,%f N\n",Pt,Ps,Pc)

-eff=Ps/P//least is Ps

+
+clc
+//solution
+//given
+t=13//mm
+ft=80//N/mm^2
+T=60//N/mm^2
+fc=120//N/mm^2
+pi=3.14
+d=6*sqrt(t)//mm//dia of rivet
+//use standard value from table 9.3
+//let p be the picth of rivets
+//Pt=(p-d)*t*ft=(p-23)*1040//N//tearing resistance of plate
+Ps=2*(pi/4)*d^2*T//N//shearing resistance of rivet
+//p-23=Ps/1040
+p=23+(Ps/1040)//mm
+//check the limits,if p&lt;=pmax..then it is safe design
+//pmax=C*t+41.28//mm=75.28mm which is more then p
+pb=0.33*p+ 0.67*d//distance btw rivets
+m=1.58*d//margin
+Pt=(p-d)*t*ft//N//tearing resistance of plate
+Ps=(2*pi/4)*d^2*T//N//shearing resistance of rivet
+Pc=2*d*t*fc//N//crushing resistance of rivet
+P=p*t*ft//N//strength of the unriveted
+//eff=(least of Pt,Ps,Pc)/P
+printf("the value of forces are,%f N\n,%f N\n,%f N\n",Pt,Ps,Pc)
+eff=Ps/P//least is Ps
 printf("the eff is,%f",eff)
\ No newline at end of file
diff --git a/278/CH9/EX9.5/ex_9_5.sce b/278/CH9/EX9.5/ex_9_5.sce
index b69a418f8..fe90624e0 100755
--- a/278/CH9/EX9.5/ex_9_5.sce
+++ b/278/CH9/EX9.5/ex_9_5.sce
@@ -1,30 +1,23 @@
-//find rivet dia,distance btw rows of rivets 

-clc

-//solution

-//given

-t=7//mm

-pi=3.14

-ft=90//N/mm^2

-T=60//N/mm^2

-fc=120//N/mm^2

-//let d be dia,since t<=8mm therefore d will be obtainned by equating shearing resistance to crushing

-//Ps=Pc

-//Ps=3*(pi/4)*d^2*T//N//shearing resistance of rivet

-//Pc=3*d*t*fc//N//crushing resistance of rivet

-//Ps=Pc

-//141.4*d^2=2520*d

-d=2520/141.4//mm

-//let p is pitch

-Ps=141.4*d^2//N

-//Pt=(p-d)*t*ft//N//tearing resistance of plate

-//Ps=Pt

-//630*(p-19)=51045

-//p=(51045/630)+19//mm

-//pmax=C*t+41.28//mm=66mm,since pmax<p..therefore p=pmax=66mm

-p=66//mm

-pb=0.33*p+0.67*d//distance btw the rivets

-Pt=(p-d)*t*ft

-Ps=141.4*d^2//N

-Pc=3*d*t*fc//N

-printf("the pitch is,%f mm\n",p)

-printf("the distance btw the rivets is,%f mm",pb)
\ No newline at end of file
+
+clc
+//solution
+//given
+t=7//mm
+pi=3.14
+ft=90//N/mm^2
+T=60//N/mm^2
+fc=120//N/mm^2
+//let d be dia,since t&lt;=8mm therefore d will be obtainned by equating shearing resistance to crushing
+//Ps=Pc
+//Ps=3*(pi/4)*d^2*T//N//shearing resistance of rivet
+//Pc=3*d*t*fc//N//crushing resistance of rivet
+//Ps=Pc
+//141.4*d^2=2520*d
+d=2520/141.4//mm
+//let p is pitch
+Ps=141.4*d^2//N
+//Pt=(p-d)*t*ft//N//tearing resistance of plate
+//Ps=Pt
+//630*(p-19)=51045
+//p=(51045/630)+19//mm
+//pmax=C*t+41.28//mm=66mm,since pmax<p..therefore p="pmax=66mm" pb="0.33*p+0.67*d//distance" btw="" the="" rivets="" pt="(p-d)*t*ft" ps="141.4*d^2//N" pc="3*d*t*fc//N" printf("the="" pitch="" is,%f="" mm\n",p)="" distance="" mm",pb)<="" div=""></p..therefore>
\ No newline at end of file
diff --git a/278/CH9/EX9.6/ex_9_6.sce b/278/CH9/EX9.6/ex_9_6.sce
index 2afa1730b..1ffd3179b 100755
--- a/278/CH9/EX9.6/ex_9_6.sce
+++ b/278/CH9/EX9.6/ex_9_6.sce
@@ -1,27 +1,27 @@
-//find rivet dia,pitch,thickness and eff

-clc

-//solution

-//given

-t=10//mm

-ft=80//N/mm^2

-T=60//N/mm^2

-pi=3.14

-//d=6*sqrt(t)//mm

-//choose standard value of d from table 9.3

-d=19//mm

-//let p is pitch of rivets

-Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet

-//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate

-//Ps=Pt

-p=19+(31900/800)//mm

-//pmax=C*t+41.28=58.78mm whihc is equal to p

-t1=0.625*t//mm

-Pt=(p-d)*t*ft//=(p-19)*800//N

-P=p*t*ft//N//strength of the unriveted

-printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)

-//eff=(least of Pt,Ps)/P

-eff=Ps/P//least is Ps

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the thickness of cover plate is,%f mm\n",t1)

+
+clc
+//solution
+//given
+t=10//mm
+ft=80//N/mm^2
+T=60//N/mm^2
+pi=3.14
+//d=6*sqrt(t)//mm
+//choose standard value of d from table 9.3
+d=19//mm
+//let p is pitch of rivets
+Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet
+//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate
+//Ps=Pt
+p=19+(31900/800)//mm
+//pmax=C*t+41.28=58.78mm whihc is equal to p
+t1=0.625*t//mm
+Pt=(p-d)*t*ft//=(p-19)*800//N
+P=p*t*ft//N//strength of the unriveted
+printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)
+//eff=(least of Pt,Ps)/P
+eff=Ps/P//least is Ps
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the thickness of cover plate is,%f mm\n",t1)
 printf("the diameter of rivets is,%f mm",d)
\ No newline at end of file
diff --git a/278/CH9/EX9.7/ex_9_6.sce b/278/CH9/EX9.7/ex_9_6.sce
index 2afa1730b..1ffd3179b 100755
--- a/278/CH9/EX9.7/ex_9_6.sce
+++ b/278/CH9/EX9.7/ex_9_6.sce
@@ -1,27 +1,27 @@
-//find rivet dia,pitch,thickness and eff

-clc

-//solution

-//given

-t=10//mm

-ft=80//N/mm^2

-T=60//N/mm^2

-pi=3.14

-//d=6*sqrt(t)//mm

-//choose standard value of d from table 9.3

-d=19//mm

-//let p is pitch of rivets

-Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet

-//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate

-//Ps=Pt

-p=19+(31900/800)//mm

-//pmax=C*t+41.28=58.78mm whihc is equal to p

-t1=0.625*t//mm

-Pt=(p-d)*t*ft//=(p-19)*800//N

-P=p*t*ft//N//strength of the unriveted

-printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)

-//eff=(least of Pt,Ps)/P

-eff=Ps/P//least is Ps

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the thickness of cover plate is,%f mm\n",t1)

+
+clc
+//solution
+//given
+t=10//mm
+ft=80//N/mm^2
+T=60//N/mm^2
+pi=3.14
+//d=6*sqrt(t)//mm
+//choose standard value of d from table 9.3
+d=19//mm
+//let p is pitch of rivets
+Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet
+//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate
+//Ps=Pt
+p=19+(31900/800)//mm
+//pmax=C*t+41.28=58.78mm whihc is equal to p
+t1=0.625*t//mm
+Pt=(p-d)*t*ft//=(p-19)*800//N
+P=p*t*ft//N//strength of the unriveted
+printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)
+//eff=(least of Pt,Ps)/P
+eff=Ps/P//least is Ps
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the thickness of cover plate is,%f mm\n",t1)
 printf("the diameter of rivets is,%f mm",d)
\ No newline at end of file
diff --git a/278/CH9/EX9.8/ex_9_6.sce b/278/CH9/EX9.8/ex_9_6.sce
index 2afa1730b..1ffd3179b 100755
--- a/278/CH9/EX9.8/ex_9_6.sce
+++ b/278/CH9/EX9.8/ex_9_6.sce
@@ -1,27 +1,27 @@
-//find rivet dia,pitch,thickness and eff

-clc

-//solution

-//given

-t=10//mm

-ft=80//N/mm^2

-T=60//N/mm^2

-pi=3.14

-//d=6*sqrt(t)//mm

-//choose standard value of d from table 9.3

-d=19//mm

-//let p is pitch of rivets

-Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet

-//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate

-//Ps=Pt

-p=19+(31900/800)//mm

-//pmax=C*t+41.28=58.78mm whihc is equal to p

-t1=0.625*t//mm

-Pt=(p-d)*t*ft//=(p-19)*800//N

-P=p*t*ft//N//strength of the unriveted

-printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)

-//eff=(least of Pt,Ps)/P

-eff=Ps/P//least is Ps

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the thickness of cover plate is,%f mm\n",t1)

+
+clc
+//solution
+//given
+t=10//mm
+ft=80//N/mm^2
+T=60//N/mm^2
+pi=3.14
+//d=6*sqrt(t)//mm
+//choose standard value of d from table 9.3
+d=19//mm
+//let p is pitch of rivets
+Ps=1*1.875*(pi/4)*d^2*T//N//shearing resistance of rivet
+//Pt=(p-d)*t*ft=(p-19)*800//N//tearing resistance of plate
+//Ps=Pt
+p=19+(31900/800)//mm
+//pmax=C*t+41.28=58.78mm whihc is equal to p
+t1=0.625*t//mm
+Pt=(p-d)*t*ft//=(p-19)*800//N
+P=p*t*ft//N//strength of the unriveted
+printf("the value of forces is,%f N\n,%f N\n",Pt,Ps)
+//eff=(least of Pt,Ps)/P
+eff=Ps/P//least is Ps
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the thickness of cover plate is,%f mm\n",t1)
 printf("the diameter of rivets is,%f mm",d)
\ No newline at end of file
diff --git a/278/CH9/EX9.9/ex_9_9.sce b/278/CH9/EX9.9/ex_9_9.sce
index 8125d075e..7c8aafe31 100755
--- a/278/CH9/EX9.9/ex_9_9.sce
+++ b/278/CH9/EX9.9/ex_9_9.sce
@@ -1,48 +1,48 @@
-//desing longitudinal joint

-clc

-//soltuion

-//given

-D=1250//mm

-P=2.5//N/mm^2

-ftu=420//N/mm^2

-fcu=650//N/mm^2

-Tu=300//N/mm^2

-eff=0.8

-Fs=5//factor of safety

-pi=3.14

-ft=ftu/Fs

-fc=fcu/Fs

-T=Tu/Fs

-t=P*D/(2*ft*eff)//mm//thickness of plate

-d=6*sqrt(t)//mm//DIA

-//Pt=(p-d)*t*ft=(p-31.5)*2100//N//tearing resistance of plate

-Ps=4*1.875*(pi/4)*d^2*T+(pi/4)*d^2*T//=8.5*(pi/4)*d^2*T//N//shearing resistance of rivet//N//shearing resistance of rivet

-//Pt=Ps

-//p-31.5=(397500/2100)

-//p=31.5+(397500/2100)//mm

-//pmax=C*t+41.28//mm=196mm

-//since p>pmax,therefore

-//p=pmax

-p=196//mm

-p'=196/2//mm

-d1=0.2*p+1.15*d//mm//diatnce between outtr and row and next row

-d2=0.165*p+0.67*d//mm//distance ebtween inner row for zigzag riveting

-t1=0.75*t//mm//thickness of wide strap

-t2=0.625*t//mm//thickness of narrow strap

-m=1.5*d//mm//margin

-Pt=(p-d)*t*ft//(p-31.5)*2100//N

-Pc=5*d*t*fc//N//crushing resistance of rivet

-P=p*t*ft//N//strength of the unriveted

-//joint may also fail due to combine teARING AND shearing reistance

-Pts=(p-2*d)*t*ft+(pi/4)*d^2*T//N

-printf("the value of forces calculted are,%f N\n,%f N\n,%f N\n",Pt,Pc,Pts)

-//eff=(least of Pt,Pc,Pts)/P

-eff=Pts/P//least is Ps

-printf("the eff is,%f\n",eff)

-printf("the pitch is,%f mm\n",p)

-printf("the thickness of wide strap is,%f mm\n",t1)

-printf("the thickness of narrow strap is,%f mm\n",t2)

-printf("the diameter of rivets is,%f mm",d)

-printf("the margine s,%f mm\n",m)

-printf("the distance btw outer and next row is,%f mm\n",d1)

+
+clc
+//soltuion
+//given
+D=1250//mm
+P=2.5//N/mm^2
+ftu=420//N/mm^2
+fcu=650//N/mm^2
+Tu=300//N/mm^2
+eff=0.8
+Fs=5//factor of safety
+pi=3.14
+ft=ftu/Fs
+fc=fcu/Fs
+T=Tu/Fs
+t=P*D/(2*ft*eff)//mm//thickness of plate
+d=6*sqrt(t)//mm//DIA
+//Pt=(p-d)*t*ft=(p-31.5)*2100//N//tearing resistance of plate
+Ps=4*1.875*(pi/4)*d^2*T+(pi/4)*d^2*T//=8.5*(pi/4)*d^2*T//N//shearing resistance of rivet//N//shearing resistance of rivet
+//Pt=Ps
+//p-31.5=(397500/2100)
+//p=31.5+(397500/2100)//mm
+//pmax=C*t+41.28//mm=196mm
+//since p&gt;pmax,therefore
+//p=pmax
+p=196//mm
+p'=196/2//mm
+d1=0.2*p+1.15*d//mm//diatnce between outtr and row and next row
+d2=0.165*p+0.67*d//mm//distance ebtween inner row for zigzag riveting
+t1=0.75*t//mm//thickness of wide strap
+t2=0.625*t//mm//thickness of narrow strap
+m=1.5*d//mm//margin
+Pt=(p-d)*t*ft//(p-31.5)*2100//N
+Pc=5*d*t*fc//N//crushing resistance of rivet
+P=p*t*ft//N//strength of the unriveted
+//joint may also fail due to combine teARING AND shearing reistance
+Pts=(p-2*d)*t*ft+(pi/4)*d^2*T//N
+printf("the value of forces calculted are,%f N\n,%f N\n,%f N\n",Pt,Pc,Pts)
+//eff=(least of Pt,Pc,Pts)/P
+eff=Pts/P//least is Ps
+printf("the eff is,%f\n",eff)
+printf("the pitch is,%f mm\n",p)
+printf("the thickness of wide strap is,%f mm\n",t1)
+printf("the thickness of narrow strap is,%f mm\n",t2)
+printf("the diameter of rivets is,%f mm",d)
+printf("the margine s,%f mm\n",m)
+printf("the distance btw outer and next row is,%f mm\n",d1)
 printf("the distance btw inner rows is,%f mm\n",d2)
\ No newline at end of file