Added new codeHEADmaster

1 files changed, 28 insertions, 0 deletions

diff --git a/3862/CH8/EX8.23/Ex8_23.sce b/3862/CH8/EX8.23/Ex8_23.sce
new file mode 100644
index 000000000..2911ca612
--- /dev/null
+++ b/3862/CH8/EX8.23/Ex8_23.sce
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+clear
+//
+
+//variable declaration
+
+// Let the force shared by bolt be Ps and that by tube be Pc. Since there is no external force, static equilibrium condition gives Ps + Pc = 0 or Ps = – Pc i.e., the two forces are equal in magnitude but opposite in nature. Obviously bolt is in tension and tube is in compression.
+//Let the magnitude of force be P. Due to quarter turn of the nut
+
+//[Note. Pitch means advancement of nut in one full turn] 
+
+Ls=(600)                      //length of whole assembly,mm
+Lc=(600)                      //length of whole assembly,mm
+delta=(0.5)
+ds=(20)                       //diameter,mm
+di=(28)                       //internal diameter,mm
+de=(40)                       //external diameter,mm
+Es=(2*100000)                 //Young's modulus, N/mm^2
+Ec=(1.2*100000)
+As=%pi*(ds**2)/4                  //area of steel bolt**mm^2
+Ac=%pi*((de**2)-(di**2))/4      //area of copper tube**mm^2
+
+P= (delta*(1/Ls))/((1/(As*Es))+(1/(Ac*Ec))) //Load,N
+
+ps=P/As                          //stress,N/mm^2
+pc=P/Ac                          //copper,N/mm^2
+
+printf("\n ps= %0.2f  N/mm^2",ps)
+printf("\n pc= %0.2f  N/mm^2",pc)