PROBLEMS ON LOCATING & CLAMPING DEVICES

Problem No. 1.7

• Q. Calculate the maximum deviation of contact pin and height of the engagement in the diamond pin locator.

From the above diagram (a) (b)

Let

E= the movement of diamond pin in the hole B

b= pin diameter in mm

D= contact pin diameter in mm

W= width across flats

H= length of pin

To find the width of contact surface between pin and hole:

D/2 = (b/2)sinα

D = b sinα

Maximum deviation of the contact pin:

From the Figure (b)

(D/2 + E/2)2 = (B/2)2 – Z2

Where Z = (b/2) cos α

(D/2) + (E/2) = [(B/2)2 – Z2]1/2

D + E = 2 [(B/2)2 – Z2]1/2

E = 2 [(B/2)2 – Z2]1/2 – D The width ‘W’ of the pin should be such that the contact surface ‘D’ is not smaller than 0.794mm. A good general rule for the dimension ‘D’ isD = B/8Length of pin,H = 3b/2If jamming of pin is avoided, length and shape of the pin must be considered.

Length/Height of EngagementThe following figure shows the limits for designing a plug to avoid sticking in a hole while the workpiece is being removed. If the plug height is,H = [2(2a+D)(D-d)]1/2

where H= height or length of pin in mma= distance from pivot to hole edgeD= diameter of holed= pin diameter of plug

The plug should not stick. ‘H’ doesn’t include chamfered portion of plug and bored hole.

Problem No. 1.8

Q. Assume a 196 mm long open end wrench is used to tighten a 18 mm bolt as shown in figure 1.127 and that a force of 41.38N is exerted at the end of the wrench. Calculate the following, if C=22.96 mm

(a) Necessary width of the clamp(b) thickness of the clamp(c) load on the bolt(d) moment on the strap(e) working stress on the clamp(f) safety factor if the ultimate stress of the clamping

material is 448.158 x 106 N/m2

(g) The maximum radial load that can be applied to this bolt

Solution:

Given, l = 196mm

d = 18.375 mm

A = 73.5 mm

B = 122.5 mm

C = 22.96 mm

F = 41.38 N

T = 41.38 x 196 x 10-3 = 8110 N-mm

(a) Width of the clamp, W = 2.3d + 1.5748 = (2.3 x 18) + 1.5748 = 42.9 mm(b) Thickness of the clamp, t = [0.85dA(1-(A/B))]1/2

= [0.85x18x73.5(1-(73.5/122.5))]1/2

= 21.2 mm (c) Load on the bolt, from T= (df)/5

so, f = 5T/d

= (5x8110)/18

= 2252.7 N

(d) The moment on the strap is,

M = fA(B-A)/B

= 2252.7 x 73.5 x 49 / 122.5

= 66238.6 N- mm

(e) The stress on the clamp is a function of section modulus of the strap.

Section modulus = (w-c)t2/6 = (42.9 – 22.96)(21.2)2/6 = 1493.6 mm3

Stress , S = M / section modulus

= 66238.6 / 1493.6

= 44.34 N/mm2

(f) Safety factor = 448.158 / 44.34

= 10.1

(g) Maximum radial force that can be placed on this clamp,

f = Sd2/ 1.352

= 44.34 x 182/1.352

= 7882.6 N

= 7.8826 KN

Problem No. 1.9• The range of holding force that may be achieved with

these clamps are from 111.209 N to 5338.078 N. These toggles may be adapted to any type of fixtures. They can be operated very quickly. Generally toggles have one fixed point and one sliding point. Find them out with the help of the diagram

The pin diameter in the figure isd = 2[(μf0)/S]1/2

wheref0= output force in NS= allowable shear stress in N/m2

μ= coefficient of frictiond= diameter of pin in mThe coefficient of friction for steel pins can be taken as μ= 0.22 for most toggles. The fixed pin is ‘a’ and the movable pin is ‘c’.