Pneumatic Gripper — Cylinder Diameter: 20-Bio-B7 Ergonomics, May 2016, Q4

Question 4: Pneumatic Gripper — Cylinder Diameter (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

The figure (PDF p. 3) shows a parallel-jaw gripper. Each jaw is carried by a parallelogram of two 3″ links pivoted on the gripper body (½″ pin spacing). The link nearer the centreline in each pair carries a slotted T-arm 1½″ along the link from its body pivot, and one pin on the ½″-diameter piston rod runs in both slots, 1¼″ perpendicular from the link line. The rod is drawn pulling to the left while the jaw arrows close, and the grip-force line is 1½″ beyond the jaw pins.

Check — stated assumptions (Note 1 of the paper): (1) pins and slots are frictionless, so the rod pin pushes on each slot normal to the slot, i.e. along the link; (2) the link angle at the gripping position is not dimensioned, so the links are taken parallel to the rod ($\theta=0$). This is the conservative case, because the required force scales with $\cos^2\theta$: at the approximately 20° drawn (the sketch is not to scale) it falls to 42.4 lbf, which needs a 0.889 in bore; (3) the ½″ dimension across the rod is its diameter, and since the rod pulls to grip, the supply pressure acts on the rod-side annulus; (4) 10 lbf is the grip force at each jaw.

Given. Grip force $G=10\ \text{lbf}$ per jaw; supply pressure $P=100\ \text{psi}$; link length $L=3\ \text{in}$; slot-pin offset $h=1.25\ \text{in}$; rod diameter $d=0.5\ \text{in}$.

Find. The pneumatic cylinder bore diameter $D$.

Fig. 4 — schematic of the source gripper, drawn with the links parallel to the rod ($\theta=0$). Each jaw rides on a 3 in parallelogram, so it translates without rotating; the rod pin drives both slotted arms at perpendicular offset $h=1.25$ in.

Approach. Because each jaw is carried by a parallelogram it translates without rotating, so the 1½″ jaw offset and the ½″ pin spacing change the link loads but not the input–output force ratio. Relate the rod and jaw velocities, apply a power (virtual-work) balance to get the rod force, then size the bore from $F=PA$ using the annulus area.

Kinematics of one jaw. If the link turns at $\omega$, the jaw pin moves at $L\omega$ perpendicular to the link, so the jaw closes at $v_{\text{jaw}}=L\omega\cos\theta$. The slot is perpendicular to the link, so the rod pin drives only its velocity component along the link: $$v_{\text{rod}}\cos\theta=h\,\omega .$$
Power balance per jaw. $F_{\text{jaw}}v_{\text{rod}}=G\,v_{\text{jaw}}$ gives $$F_{\text{jaw}}=\frac{G\,L\cos^2\theta}{h}=\frac{10(3)(1)}{1.25}=24\ \text{lbf}.$$ The linkage reduces force: $G/F_{\text{jaw}}=h/L=1.25/3=0.417$, because the 3 in output arm is longer than the 1.25 in input arm.
Total rod force. The one rod pin drives both jaws: $$F_{\text{cyl}}=2F_{\text{jaw}}=\boxed{48\ \text{lbf}}.$$
Required effective area. $$A=\frac{F_{\text{cyl}}}{P}=\frac{48}{100}=0.48\ \text{in}^2.$$
Bore diameter (rod-side annulus). $\frac{\pi}{4}(D^2-d^2)=A$, so $$D=\sqrt{\frac{4A}{\pi}+d^2}=\sqrt{\frac{4(0.48)}{\pi}+0.5^2}=\sqrt{0.611+0.250}=\boxed{0.928\ \text{in}\ (23.6\ \text{mm})}.$$ Select the next standard size, a 1 in bore, which pulls with $100\cdot\frac{\pi}{4}(1^2-0.5^2)=58.9\ \text{lbf}\ge 48\ \text{lbf}$. Ignoring the rod (full piston area) would give 0.782 in, which would be undersized on the pull stroke.

An independent free-body solution of one jaw and its two links reproduces the virtual-work rod force exactly.

Result	Value
Force ratio per jaw $G/F_{\text{jaw}}=h/L$	0.417
Rod force per jaw	24 lbf
Cylinder force $F_{\text{cyl}}$	48 lbf
Required annulus area	0.48 in$^2$
Minimum bore $D$	0.928 in (23.6 mm); use a 1 in standard bore
Sensitivity: links at the drawn approx. 20°	42.4 lbf; 0.889 in bore

Concept & Theory

Topic. Force transmission through a pin-in-slot parallelogram gripper linkage, and pneumatic cylinder sizing.

Key relations. Virtual work $F_{\text{in}}v_{\text{in}}=F_{\text{out}}v_{\text{out}}$; per jaw $F_{\text{jaw}}h=G\,L\cos^2\theta$; pull-stroke force $F=P\cdot\frac{\pi}{4}(D^2-d^2)$.

Why this works. In an ideal (frictionless) mechanism the input power equals the output power, so the force ratio is the inverse of the velocity ratio. A parallelogram makes the jaw translate, so only the link length $L$ sets the jaw's speed, and the slot offset $h$ sets how fast the rod turns the link.

Common pitfall. Writing the lever ratio upside down. From $F_{\text{in}}r_{\text{in}}=F_{\text{out}}r_{\text{out}}$, a long output arm reduces the output force, so $MA=r_{\text{in}}/r_{\text{out}}$, not $r_{\text{out}}/r_{\text{in}}$. A second pitfall is sizing from the full piston area when the rod pulls to grip.

Source. Groover, Weiss, Nagel & Odrey, Industrial Robotics: Technology, Programming, and Applications, ch. on end-effectors and grippers.

Question 4 of 6: Pneumatic Gripper — Cylinder Diameter

Question 4: Pneumatic Gripper — Cylinder Diameter (20 marks)