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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /278/CH17/EX17.16/ex_17_16.sce | |
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initial commit / add all books
Diffstat (limited to '278/CH17/EX17.16/ex_17_16.sce')
-rwxr-xr-x | 278/CH17/EX17.16/ex_17_16.sce | 69 |
1 files changed, 69 insertions, 0 deletions
diff --git a/278/CH17/EX17.16/ex_17_16.sce b/278/CH17/EX17.16/ex_17_16.sce new file mode 100755 index 000000000..965c23ab8 --- /dev/null +++ b/278/CH17/EX17.16/ex_17_16.sce @@ -0,0 +1,69 @@ +//desing
+clc
+//soltuion
+//given
+//ref fig 17.12
+W=4000//N
+l=110//mm
+ft=100//N/mm^2
+t=50//N/mm^2
+pb=20//N/mm^2
+p=6
+u=0.20
+//design of square
+//cos(q)=0.8112
+q=(%pi/180)*35.1//rad
+F=W/(2*tan(q))//N
+W1=2*F//N
+//let dc be core dia
+//dc=sqrt(4*W1/(%pi*ft))
+printf("the core dia is,%f mm\n",sqrt(4*W1/(%pi*ft)))
+printf("the core dia is,say dc=14 mm\n")
+dc=14//mm
+do=dc+p
+printf("the nominal dia is,%f mm\n",do)
+d=do-p/2
+//tan(a)=p/(%pi/d)=b=0.1123
+b=0.1123
+P=W*[(b+u)/(1-(b*u))]//N
+T=(P*d/2)
+t1=(16*T)/(%pi*dc^3)///N/mm^2
+ft1=W1/(%pi/4*dc^2)//N/mm^2
+ftmax=0.5*(ft1+ sqrt(ft1^2 +4*t1^2))//N/mm^2
+tmax=0.5*(sqrt(ft1^2 + 4* t1^2))//N/mm^2
+printf("the max prin stress and max shear stress is,%f N/mm^2\n,%f N/mm^2\n",ftmax,tmax)
+printf("since max stresses are within safe limits,therefore design is safe\n")
+//design of nut
+//let n be numbr of threads
+//n=W1/(%pi/4*20*(do^2-dc^2))
+printf("the numbr of threads are ,%f \n",W1/(%pi/4*20*(do^2-dc^2)))
+printf("since number of threads can not so less,so we take n=4\n")
+n=4
+t2=n*p//mm
+printf("the number of therd and thickness of threads is,%f \n,%f mm\n",n,t2)
+b1=1.5*do
+printf("the width of nut is,%f mm\n",b1)
+length =210+t2+(2*8)
+printf("the length of screwd portion is,%f mm\n",length)
+//desig of pins in nuts
+//let d1 be dia
+d1=sqrt(F/(2*%pi/4*t))
+printf("the dia of pins in nuts is,%f mm\n",d1)
+//design of links
+F1=F/2//load on link
+Wcr=1423*5//Fs=5
+//let t3 be thickness and b3 be width of link
+//Al=t3*3t3=3*t3^2//b3=3*t3
+//I=(1/12)*t3*b3^2=2.25*t3^4
+//k=sqrt(I/Al)=0.866*t3
+L=110
+a=1/7500
+//acc tor ankine fornula
+//Wcr=(ft*Al)/(1+a*(L/k)^2)=300*t3^2/(1+(2.15/t1^2))
+//t3^4-23.7*t3^2-51=0
+//t3=sqrt(25.7)//mm
+printf("the thickness and width is,%f mm\n,%f mm\n",sqrt(25.7),3*sqrt(25.7))
+printf("the thickness is say 6mm\n")
+t3=6//mm
+b3=3*t3//mm
+Al=3*t3^2
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