Question 1.CS.1: Bolt Cutter Loading Analysis Many components, such as bicycl......
Bolt Cutter Loading Analysis
Many components, such as bicycle levers, automotive scissors jacks, bolt cutting tools, various types of pliers, and pin-connected symmetrical assemblies, may be treated by applying Equation (1.5), similar to that which will be illustrated here. We note that a mechanical linkage system is designed to transform a given input force and movement into a desired output force and movement. In this case, accelerations on moving bars require that a dynamic analysis be done through the use of Equation (1.7). Bolt cutters can be used for cutting rods (see Section I section opener page), wire mesh, and bolts. Often, a bolt cutter’s slim cutting head permits cutting close to surfaces and incorporates one-step internal cam mechanism to maintain precise jaw or blade alignment. Handle design and handle grips lend to controlled cutting action. Jaws are manufactured from heat-treated, hardened alloy steel.
\begin{array}{cccc} \Sigma F_x=0 & \Sigma F_y=0 & \Sigma F_z=0 \\ \Sigma M_x=0 & \Sigma M_y=0 & \Sigma M_z=0 \end{array} (1.5)
\Sigma F_x=m a_x \quad \Sigma F_y=m a_y \quad \Sigma M_z=I \alpha (1.7)
Figure 1.4 depicts schematic drawing of a bolt cutter, a pin-connected tool in the closed position in the process of gripping its jaws into a bolt. The user provides the input loads between the handles, indicated as the reaction pairs P . Determine the force exerted on the bolt and the pins at joints A, B , and C .
Given
The geometry is known. The data are
P=2 lb , \quad a=1 \text { in., } \quad b=3 \text { in., } \quad c=\frac{1}{2} \text { in., } \quad d=8 \text { in., } e=1 \text { in., }
Assumptions
Friction forces in the pin joints are omitted. All forces are coplanar, 2D, and static. The weights of members are neglected as being insignificant compared to the applied forces.
The equilibrium conditions are fulfilled by the entire cutter. Let the force between the bolt and the jaw be Q , whose direction is taken to be normal to the surface at contact (point D ). Due to the symmetry, only two FBDs shown in Figure 1.5 need to be considered. Inasmuch as link 3 is a two-force member, the orientation of force F_A is known. Note also that the force components on the two elements at joint B must be equal and opposite, as shown on the diagrams.
Conditions of equilibrium are applied to Figure 1.5a to give F_{Bx}=0 and
\Sigma F_y=Q-F_A+F_{B y}=0 \quad F_A=Q+F_{B y}
\Sigma M_B=Q(4)-F_A(3)=0 \quad F_A=\frac{4 Q}{3}
from which Q=3F_{By} . In a like manner, referring to Figure 1.5b, we obtain
\Sigma F_y=-F_{B y}+F_{C y}-2=0 \quad F_{C y}=\frac{Q}{3}+2
\Sigma M_C=F_{B y}(1)+F_{B y}(0.5)-2(8)=0 \quad F_{B y}=32 lb
and F_{Cx} =0. Solving Q =3(32)=96 lb The shear forces on the pins at the joints A, B , and C are
F_A=128 lb , \quad F_B=F_{B y}=32 lb , \quad F_C=F_{C y}=34 lb
Comments: Observe that the high mechanical advantage of the tool transforms the applied load to a large force exerted on the bolt at point D . The handles and jaws are under combined bending and shear forces. Stresses and deflections of the members are taken up in Case Studies 3.1 and 4.1 in Chapters 3 and 4, respectively. The MATLAB solution of this case study and some others are on the website (see Appendix E).
