Example 1a-9. The Bell Crank, Which Is in Equilibrium Under The Forces Shown in The Figure, Is

 

Example 1a-9. The bell crank, which is in equilibrium under the forces shown in the igure, is supported by a 20-mm-diameter pin at D.

Determine: a. the required diameter of the connecting rod AB, given that that its tensile working str stress ess is 100 MPa; and b. the shear str stress ess in the pin.

Pin at D

Solution: For a:

Step 1: Since the problem asks for the required diameter of the connecting rod AB, given that  the working stress is 100 MPa. Then we must determine the value of  P AB by statics:

CCW+ ¿  ; ∑ M  D= 0 ; CCW+

 P AB ( 200 )=30si 30sin n 60 ° ( 240 )  P AB=31.177kN  P AB

100

If σ allowable ≤

100

31. 177 kN 

 N  mm

MPa → 

2

=

π 

 

(

σ = A  AB

 )

1000 N  1 kN  2

For b:

Step 1: Since the problem problem asks for the shear stress stress in the pin, at D. Therefore Therefore,, we must irst  determine the reaction at D by statics:

 

∑ F  y =0  ; ↑ +¿ ;

 D y −30si 30sin n 60 ° = 0  D y = 25.981 kN

∑ F  x =0 ; → + ¿ ;

 D x −30cos60 ° − P AB= 0   D x −30cos60 ° −31.177 =0  D x =46.177 kN

2 2 D X  ) = √ 25.981 25.981 + 46.177 =¿  52.984 kN  R D= √ ( D  y ) + (  D 2

2

Step 2: Determine whether single single   or double double shear.  shear. Looking at the connection detail, it can be concluded that the connection is indeed a double shear. Now we can substitute it into the formula: With  D pin =20 mm and  R D= ¿ 52.984 kN and (double shear)

 R D 2   V  τ = = =  A  pin  A  pin

52.984 2

π  4

kN x

(

 )

1000 N  1 kN 

( 20 ) mm 2

2

Example 1a-10.  Compute Compute the maximum force P that can be applied applied to the foot pedal. The 6mm.-diameter pin at B is in single shear, and its working shear stress is 28 MPa. The cable has a diameter of 3 mm attached at pin C and the cable has also a working normal stress of 140 MPa.

 

Solution:

Step 1. Analyze the problem. Since the given in the problem are allowable stresses at pin B and at the cable, we start there. Based on the working stresses, we can determine the capacity of the pin  R B and tension at the cable, T . At pin: (the problem mentioned single single shear):  shear): Given :  D pin =6 mm and τ allowable ≤ 28MPa =

V 

=

τ   A

 R B

=

 Pcable

 D pin =3 mm and σ allowable ≤ 140MPa

  T 

=

 A  pin σ   A cable

 A cable

 R B N 

  N 

  28

  N 

= 2

mm

π  4

140

( 6 ) mm 2

2

2

mm

=

  T N  π  2 ( 3 ) mm 2 4

Step 2. But the problem asks for the maximum force P that can be applied in the system. Now we shall solve the forces  R B  and T in terms of what was asked in the system, which is P, through statics: