Ex-2-5a Calculations
Set variables to be non-zero and constants to be positive in order to avoid possible
zero denominators in the symbolic equations.
assume(T1<>0 and T2<>0 and d1<>0 and d2<>0 and aB<>0 and ac<>0 and E1>0 and E2>0 and T1a<>0 and T2a<>0):
asume(sin(aB)<>0 and sin(aC)<>0 and a>0 and b>0 and P<>0 and P1<>0 and P2<>0 and d3<>0 and d4<>0):
assume(L1>0 and L2>0 and A1>0 and A2>0 and k1<>0 and k2<>0 and P<>0 and s1<>0 and s2<>0):
assume(P1a<>0 and P2a<>0):
The equilibrium equation for T1 and T2
f1:=T1*b*sin(aB)+T2*2*b*sin(aC)-2*P*b=0
T1 b sin aB - 2 P b + 2 T2 b sin aC = 0
The compability equation for δ1 and δ2
d2:=2*(sin(aC)/sin(aB))*d1
2 d1 sin aC
sin aB
The force-displacement equations for δ1 and δ2, T1 and T2 are (let k1=(L1/E1*A1 and k2=(L2/E2*A2))
d1:=k1*T1
T1 k1
Note: d3 is just a different name for d2 so the program variables don't get confused.
d3:=k2*T2
T2 k2
Solving for δ2 =(d2 in the equation below)
solve([d2=2*sin(aC)/sin(aB)*d1,d1=k1*T1],[d2])
2 T1 k1 sin aC = z
sin aB


  

   


  
   


  
  
  
  
   




The expression for δ2 as a function of T1 is
d2:=(2*T1*k1*sin(aC))/sin(aB)
2 T1 k1 sin aC
sin aB
Solving for T2 as a function of T1 since δ2 = k2*T2 also
solve([d2=d3],[T2])
T2 = 2 T1 k1 sin aC
k2 sin aB
T2:=(sin(aC)/sin(aB))*(k1/k2)*2*T1
2 T1 k1 sin aC
k2 sin aB
The equilibrium equation, ΣA is
MA:=T1*b*sin(aB)+T2*2*b*sin(aC)-P*2*b=0
4 T1 b k1 sin aC 2
T1 b sin aB - 2 P b +
=0
k2 sin aB
Solving for T1 using the above expressions for MA andT2
solve(T1*b*sin(aB)+T2*2*b*sin(aC)-P*2*b,T1)
C
0
if b = 0
2 P k2 sin aB
s1
if s1 ¹ 0 Ù b ¹ 0 Ù aB
ÏZ
p
Æ
if s1 = 0 Ù b ¹ 0 Ú
aB
p
ÎZ
where
s1 = k2 sin aB 2 + 4 k1 sin aC


2
Factoring out the term, k2, a term (k1/k2) emerges as an important coefficient
T1:=2*P*sin(aB)/(sin(aB)^2 + 4*(k1/k2)*sin(aC)^2)
Going back to the equation for T2 and subsituting the above equation for T1
T2:=(sin(aC)/sin(aB))*(k1/k2)*2*2*P*sin(aB)/(sin(aB)^2 +(k1/k2)*4*sin(aC)^2)
4 P k1 sin aC
   

   
k2 sin aB 2 +
4 k1 sin aC 2
k2
If P1 is the allowable load for wire BD with tensile force, T1a (a dummy variable for T1), and the allowable stress is σ1=T1a/A1, then, solving for P1
solve([T1a=2*P1*sin(aB)/(sin(aB)^2 + 4*(k1/k2)*sin(aC)^2)],[P1])
solve([T1a=2*P1*sin(aB)/(sin(aB)^2 + 4*(k1/k2)*sin(aC)^2)],[P1])
       
 
 
     


 
      

  
  

  

     


4 k1 sin aC
T1a sin aB 2 +
k2
2 sin aB
2
=z
if T1a = A1 s1
Æ
if T1a ¹ A1 s1
The expression for P1 is (s1=σ1 in the equation below)
P1:=s1*A1*(sin(aB)^2+4*(k1/k2)*sin(aC)^2)/(2*sin(aB))
4 k1 sin aC 2
k2
A1 s1 sin aB 2 +
2 sin aB
If P2 is the allowable load for wire CD with tensile force, T2a (a dummy variable for T2), and the allowable stress is σ2=T2/A2, then solving for P2
solve([T2a=(4*P2*(k1/k2)*sin(aC)/(sin(aB)^2+4*(k1/k2)*sin(aC)^2))],[P2])
P2 =
T2a k2 sin aB 2 + 4 T2a k1 sin aC 2
4 k1 sin aC
if k1 ¹ 0 Ù aC
ÏZ
p
P2 = z
if k1 = 0 Ú
aC
p
Î Z Ù T2a = 0
Æ
if k1 = 0 Ú
aC
Î Z Ù T2a ¹ 0
p
P2:=s2*A2*((k2/k1)*sin(aB)^2+4*sin(aC)^2)/(4*sin(aC))
A2 s2 sin aB 2 +
4 k1 sin aC 2
k2
4 sin aC
Ex 2-5 Calculations
Using the above expressions, the values for P1 and P2 can now be found. Let P1a and P2a be dummy variables for P1 and P2.
Calculating the wire areas for BD and CD
A1:=(3.1416/4)*(4.2*unit::mm)^2
13.854456 mm2
A1=13.86 mm^2
A2:=(3.1416/4)*(3.2*unit::mm)^2
8.042496 mm2
A2= 8.04 mm^2
Calculating the lengths of wires BD and CD
L1:=sqrt((1.8*unit::m)^2+(1.2*unit::m)^2)*1000*unit::mm/unit::m
2163.330765 mm
L1=2.16 m
L2:=sqrt((1.8*unit::m)^2+(2.4*unit::m)^2)*1000*unit::mm/unit::m
3000.0 mm
L2=3.00 m
Calculating the terms, k1=L1/(E1*A1) and k2=L2/(E2*A2)
k1:=L1/(72*unit::GPa/(unit::GPa/(10^3*unit::N/unit::mm^2))*A1)
0.002168707355 mm
N
k1=2,17*10^-3 mm
k2:=L2/(45*unit::GPa/(unit::GPa/(10^3*unit::N/unit::mm^2))*A2)
0.008289300569 mm
N
k2=8.29*10^-3 mm


Calculating aB
aB:=arctan(1.8/1.2)
0.9827937232
aB=0.983 rad
Calculating aC
aC:=arctan(1.8/(2*1.2))
0.6435011088
aC= .644 rad
Let P1a and P2a be the dummy variables for P1 and P2
s1:=200*unit::MPa/(unit::MPa/(unit::N/unit::mm^2))
200 N 2
mm
P1a:=s1*A1*(sin(aB)^2+4*(k1/k2)*sin(aC)^2)/(2*sin(aB))
1780.075107 N
P1a=1780 N

P1a=1780 N
s2:=172*unit::MPa/(unit::MPa/(unit::N/unit::mm^2))
172 N 2
mm
P2a:=s2*A2*((k2/k1)*sin(aB)^2+4*sin(aC)^2)/(4*sin(aC))
2355.176296 N

P2a=2355 N
The final answer is P = min(P1a,P2a)=1780 N