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author | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
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committer | prashantsinalkar | 2018-02-03 11:01:52 +0530 |
commit | 7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch) | |
tree | 449d555969bfd7befe906877abab098c6e63a0e8 /3864/CH8/EX8.14/Ex8_14.sce | |
parent | d1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff) | |
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Diffstat (limited to '3864/CH8/EX8.14/Ex8_14.sce')
-rw-r--r-- | 3864/CH8/EX8.14/Ex8_14.sce | 73 |
1 files changed, 73 insertions, 0 deletions
diff --git a/3864/CH8/EX8.14/Ex8_14.sce b/3864/CH8/EX8.14/Ex8_14.sce new file mode 100644 index 000000000..b923e1abc --- /dev/null +++ b/3864/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,73 @@ +clear +// +// + +//Initilization of Variables + +d_o=300 //mm //Outside diameter +d1=180 //mm //Internal Diameter +p=12 //N/mm**2 //internal Fluid pressure +p_o=6 //N/mm**2 //External Pressure +r_o=150 //mm //Outside Diameter +r=90 //mm //Internal Diameter + + +//Calculations + +//From Lame's Equation we have +//p_x=b*(x**2)**-1-a ..........................(1) +//F_x=b*(x**2)**-1+a ...........................(2) + +//At +x=90 //N/mm**2 +r1=90 //N/mm**2 +p=42 //N/mm**2 +//Sub in equation 1 we get +//42=b*(90**2)**-1-a ..............................(3) + +//At +p2=6 //N/mm**2 +//sub in equation 1 we get +//6=b*(150**2)**-1-a ..............................(4) + +//From equations 3 and 4 weget +//36=b*(90**2)**-1-b2(150**2)**-1 +//After further simplifying we get +b=36*90**2*150**2*(150**2-90**2)**-1 + +//Sub value of b in equation 4 we get +a=b*(150**2)**-1-p_o + +//At +F_x=b*(x**2)**-1+a //N/mm**2 + +//At +x2=150 //mm +r=150 //mm + +F_x2=b*(x2**2)**-1+a //N/mm**2 + +//Now if External pressure is doubled i.e p_o2=12 //N/mm**2 We have +p_o2=12 //N/mm**2 +//sub in equation 4 we get +//12=b2*(150**2)**-1-a2 ..........................(5) + +//Max Hoop stress is to be 70.5 //N/mm**2,which occurs at x=r1=90 //mm +//Sub in equation 4 we get +//70.5=b*(90**2)**-1+a2 ................................(6) + +//Adding equation 5 and 6 +//82.5=b2*(150**2)**-1+b*(90**2)**-1 +//After furhter simplifying we get +b2=82.5*150**2*90**2*(150**2+90**2)**-1 + +//Sub in equation 5 we get +a2=b2*(150**2)**-1-12 + +//If p_i is the internal pressure required then from Lame's theorem +p_i=b2*(r1**2)**-1-a2 + +//Result +printf("\n Stresses int the material are:F_x %0.2f N/mm**2",F_x) +printf("\n :F_x2 %0.2f N/mm**2",F_x2) +printf("\n Internal Pressure that can be maintained is %0.2f N/mm**2",p_i) |