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Diffstat (limited to '3864/CH2/EX2.20/Ex2_20.sce')
| -rw-r--r-- | 3864/CH2/EX2.20/Ex2_20.sce | 56 |
1 files changed, 56 insertions, 0 deletions
diff --git a/3864/CH2/EX2.20/Ex2_20.sce b/3864/CH2/EX2.20/Ex2_20.sce new file mode 100644 index 000000000..9d5785310 --- /dev/null +++ b/3864/CH2/EX2.20/Ex2_20.sce @@ -0,0 +1,56 @@ +clear +// +// + +//Initilization of Variables + +sigma=150 //N/mm**2 //Stress +P=40*10**3 //N //Load + +//Calculations + +//LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively + +//Let sum of all Vertical Forces=0 +//P_A+P_B+P_C+P_D=40 ..........................(1) + +//Let x be the distance between each wires +//sum of all moments=0 +//P_B*x+P_C*2*x+P_D*3*x=40*2*x + +//After further simplifying we get +//P_B+2*P_C+3*P_D=80 ..........................(2) + +//As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations + +//Let dell_l be the increse in elongation of wire + +//dell_l_B=dell_l_A+dell_l +//dell_l_C=dell_l_A+2*dell_l +//dell_l_D=dell_l_A+3*dell_l + +//Let P1 be the force required for the Elongation of wires,then +//P_B=P_A+P1 ] +//P_C=P_A+2*P1 ] +//P_D=P_A+3*P1 ] ................................(3) + +//from Equation (3) and (1) we get +//2*P_A+3*P1=20 ................................(4) + +//from Equation (3) and (2) we get +//6*P_A+14*P1=80 + +//subtracting 3 times equation (4) from (3) we get +P1=20*5**-1 + +//from Equation 4 we get +P_A=(80-14*P1)*6**-1 +P_B=P_A+P1 +P_C=P_A+2*P1 +P_D=P_A+3*P1 + +//Let d be the diameter required,then +d=(P_D*10**3*4*(%pi*150)**-1)**0.5 + +//result +printf("\n The Required Diameter is %0.2f mm",d) |