A quick return mechanism of the crank and slotted lever type shaping machine is shown in Fig. 7.17.
The dimensions of the various links are as follows :
$O_{1} O_{2}=800 mm ; O _{1} B =300 mm$ ;
$O_{2} D=1300 mm ; D R=400 mm$ .
The crank $O_{1} B$ makes an angle of $45^{\circ}$ with the vertical and rotates at 40 r.p.m. in the counter clockwise direction. Find : 1. velocity of the ram R, or the velocity of the cutting tool, and 2. angular velocity of link $O_{2} D$ .

Question

A quick return mechanism of the crank and slotted lever type shaping machine is shown in Fig. 7.17.
The dimensions of the various links are as follows :
[latex]O_{1} O_{2}=800 mm ; O _{1} B =300 mm[/latex];
[latex]O_{2} D=1300 mm ; D R=400 mm[/latex].
The crank [latex]O_{1} B[/latex] makes an angle of [latex]45^{\circ}[/latex] with the vertical and rotates at 40 r.p.m. in the counter clockwise direction. Find : 1. velocity of the ram R, or the velocity of the cutting tool, and 2. angular velocity of link [latex]O_{2} D[/latex].

Accepted Answer

Given: [latex]N_{ BO 1}=40 ext { r.p.m. or } \omega_{ BO 1}=[/latex][latex]2 \pi imes 40 / 60=4.2 rad / s[/latex]

Since the length of crank [latex]O_{1} B=300 mm =0.3 m[/latex],therefore velocity of B with respect to [latex]O_{1}[/latex] or simply velocity of B (because [latex]O_{1}[/latex] is a fixed point),

[latex]v_{ BO 1}=v_{ B }=\omega_{ BO 1} imes O_{1} B=4.2 imes 0.3=1.26 m / s[/latex] [latex]\ldots ext { (Perpendicular to } \left.O_{1} B ight)[/latex]

1. Velocity of the ram R

First of all draw the space diagram, to some suitable scale, as shown in Fig.(a). Now the velocity diagram, as shown in Fig.(b), is drawn as discussed below :

1. Since [latex]O_{1} ext { and } O_{2}[/latex] are fixed points, therefore these points are marked as one point in the velocity diagram. Draw vector [latex]o_{1} b[/latex] perpendicular to [latex]O_{1} b[/latex], to some suitable scale, to represent the velocity of B with respect to [latex]O_{1}[/latex] or simply velocity of B, such that

[latex] ext { vector } o_{1} b=v_{ BOl }=v_{ B }=1.26 m / s[/latex]

2. From point [latex]o_{2} [/latex], draw vector [latex]o_{2} c[/latex] perpendicular to [latex]O_{2} c[/latex] to represent the velocity of the coincident point C with respect to [latex]O_{2}[/latex] or simply velocity of C (i.e. [latex]v_{ CO 2} ext { or } v_{ C }[/latex] ), and from point b, draw vector bc parallel to the path of motion of the sliding block (which is along the link [latex]O _{2} D[/latex]) to represent the velocity of C with respect to B (i.e. [latex]v_{ CB }[/latex]). The vectors [latex]O _{2} c[/latex] and bc intersect at c.

3. Since the point D lies on [latex]O_{2} C[/latex] produced, therefore divide the vector [latex]o_{2} c[/latex]at d in the same ratio as D divides [latex]O_{2} C[/latex] in the space diagram. In other words,

[latex]cd / o _{2} d = CD / O _{2} D[/latex]

4. Now from point d, draw vector dr perpendicular to DR to represent the velocity of R with respect to D (i.e. [latex]v_{ RD }[/latex] ), and from point [latex]o_{1}[/latex] draw vector [latex]O_{1} r[/latex] parallel to the path of motion of R (which is horizontal) to represent the velocity of R (i.e.[latex]v_{ R }[/latex]). The vectors dr and [latex]o_{1} r[/latex] intersect at r.

By measurement, we find that velocity of the ram R,

[latex]v_{ R }= ext { vector } o_{1} r=1.44 m / s[/latex]

2. Angular velocity of link [latex]O_{2} D[/latex]

By measurement from velocity diagram, we find that velocity of D with respect to [latex] O _{2}[/latex] or velocity of D,

[latex]v_{ DO 2}=v_{ D }= ext { vector } o_{2} d=1.32 m / s[/latex]

We know that length of link [latex]O_{2} D[/latex] = 1300 mm = 1.3 m. Therefore angular velocity of the link [latex]O_{2} D[/latex] ,

[latex]\omega_{ DO _{2}}=\frac{v_{ DO 2}}{O_{2} D}=\frac{1.32}{1.3}=1.015 rad / s ext { (Anticlockwise about } \left.O_{2} ight)[/latex]

Question : A quick return mechanism of the crank and slotted lever type...