Example 2

In this second example the same centric slider-crank mechanism of the previous example is considered. However, in addition to F₁₄ two additional forces, F₁₃ and F₁₂, act on the links. F₁₂ = 100 N , F₁₃ = 150 N and F₁₄ = 100 N.

The free body diagrams of the links are shown below. The pin radii and the friction coefficient are that of the previous example.

The force equilibrium equations for link 4:

SFx = F34x − F14 − μG14=0       or       F34x = 100 + 0.1G14	(i)
SFy = F34y − G14 = 0       or       G14 = F34y	(ii)
SMB = −sG14 − 5G14 + 5F34 = 0      or       s = 5(F34 − G14)/G14	(iii)

If s is within the physical boundaries of the two contacting surfaces, the above equations are valid (contact Mode I or II will exist). Also:

\displaystyle {\text{F}}_{\text{34}}=\sqrt{{{{\text{F}}_{\text{34x}}}^2+{{\text{F}}_{\text{34y}}}^2}}

(iv)

For link 3:

SFx = −F34x + F13cos(210°) + F23x = 0       or       F23x = F34x + 129.904	(v)
SFy = F34y + F13sin(210°) + F23y = 0       or      F23y = 75 − F34y	(vi)
SMB = −a3F34ycosθ13 − a3F34xsinθ13 − 25F23 − 5F34 − 400F13sin(210° − θ13) = 0      or 783.045F34y − 163.829F34x − 25F23 − 5F34 − 40005.218 = 0	(vii)

and

\displaystyle {\text{F}}_{\text{23}}=\sqrt{{{{\text{F}}_{\text{23x}}}^2+{{\text{F}}_{\text{23y}}}^2}}

(viii)

For link 2:

SFx = −F23x + G12x + F12cos(−60°) = 0       or       G12x = F23x − 50	(ix)
SFy = F23y + G12y + F12sin(−60°) = 0       or      G12y = F23x + 86.603	(x)
SMA0 = T12 + 100F12sin(−60° − θ12) + 200F23ysin(θ12) − 200F23xcos(θ12) − 25F23 − 5G12 = 0      or T12 = 9063.078 − 163.830F23x + 114.715F23y +25F23 + 5G12	(xi)

and

\displaystyle {\text{G}}_{\text{12}}=\sqrt{{{{\text{G}}_{\text{12x}}}^2+{{\text{G}}_{\text{12y}}}^2}}

(xii)

In the equations the forces are in N (Newton) and moments are in N·mm. There are 12 equations in 12 unknowns (F_34x , F_34y , F₃₄, F_23x, F_23y, F₂₃, G_12x , G_12y, G₁₂ , G₁₄ , s and T₁₂).   Equations (iv, viii, xii) are non-linear, whereas the remaining equations are linear. Direct solution of these equations will not be attempted. An iterative procedure is more suitable in such cases.

Substituting Equations (i) and (ii) into equation (vii) and solving for G₁₄ yields:

G14 = 265.872 + 0.033F23 + 0.007F34

(xiii)

The last two terms of equation (xiii) are due to friction at the joints and it is logical that these terms will be small in comparison to the first term, which is the value of G₁₄ when μ = 0. As the first guess for G₁₄, if we neglect the effect of friction, the value of G₁₄ can be easily determined (G₁₄ = 265.872 N). We can than solve for F_34x using equation (i), F_34y and F₃₄ from equation (ii), F_34y = G₁₄ then use equation (iv). Next using equations (v, vi, viii)  we can solve for F_23x , F_23y , F₂₃ . The values of F₂₃ and F₃₄ thus found can now be used to refine the value of G₁₄. The procedure is repeated until there is no significant change in G₁₄. The iteration steps for this example are shown in the table below. The procedure in general rapidly converges, since the friction effect in mechanisms is quite small (if the mechanism is not at a critical position). Also one can see the effect of friction on the solution.

Force                                   2. Step       3. Step      4. Step         5. Step

G₁₄(eq. xiii) (G₁₄⁰ = F_34y)    265.872      278.484     278.854         278.865

F_34x (eq. iv)                        126.587      127.848      127.885        127.886

F_34y (eq. iv)                         294.469     306.429      306.780        306.791

F_23x (eq. v)                          256.491     257.752      257.789        257.790

F_23y (eq. vi)                        -190.872    -203.484     -203.854       -203.865

F₂₃  (eq. vii)                         319.718      328.393      328.652       328.659

(All values in Newton)

After this iterative method of finding F₃₄, G₁₄ and F₂₃, one can than solve for the other unknowns from the remaining equations:

s = 0.96 mm  (eq. iii),     G_12x = 207.790 N  (eq. ix),     G_12y = −117.262 N (eq. x)     and     G₁₂ = 238.594 N (eq. xii)

T₁₂ = 47.148 N·m (CW)

Since the friction effect is usually small, the above iterative method will converge to a solution in very few steps.