In proportioning pin-connected joints, the shear of the pin and the bearing value of the connected plates should be considered in the same manner as was riveted joints. Pins, however, are subjected to bending stresses that are neglected in riveted work; pins should always be proportioned to withstand such bending stresses. Round pins should be considered as beams having a solid circular cross-section. The bending moment in inch-pounds should be determined in a similar manner as for any beam (see pages 107 to 120), and the resisting moment of the section obtained, which may be calculated (see page 114), or obtained from Table XXX, which also gives the shearing and bearing values for different size pins.


What size pin will be required to resist bending in the connection shown in Fig. 32?


The bending moment is 10,000 lb. X 6 in. = 60,000 in.-lb.

The proper size pin having the required resisting moment may be obtained from Table XXX, page 137, or calculated thus: The section modulus for a solid cylindrical section is, accord-ing to Table XII, page 83, .0982 d3. Since d, the diameter of pin, must be assumed, try 3 1/2 in. The section modulus is .0982 X3 1/2X3 1/2X3 1/2, or 4.21. Then, as the safe resisting moment of any beam equals the section modulus multiplied by the safe stress of the material page 114), and assuming that the safe working stress of the material in the pin is 15,000 lb. per sq. in., the resisting moment is 4.21 X 15,000 = 63,150 in.-lb. Since the safe resisting moment must equal or exceed the bending moment (sea page 108), a pin 3 1/2 in. in diameter will be sufficient to resist the bending stresses.

Where the lines of action of the stresses, in several members connected at a common joint by a pin, are inclined to one another, as at (a), Fig. 33, the stresses in the oblique members should be resolved into their vertical and horizontal components (see page 141). Having found these for all the forces acting upon the pin, the greatest bending moment due to all the vertical components, and that due to the horizontal components, should be obtained. Then, by adding the squares of these two amounts together and taking the square root of the result, the greatest resultant bending moment will be found.

Strength Of Pins 269

Fig. 32.

Strength Of Pins 270


Find the greatest resultant bending moment upon the pin shown at (a), Fig. 33.


Draw accurately the frame diagram, as at (6), Fig. 33, in which the full lines represent the direction of the members assembled at the joint. Lay off to some convenient scale upon c d, a distance that will represent the stress in that member; .then, by the principle of the parallelogram of forces, db will be the horizontal, and de the vertical, component the stress in cd. Likewise, draw the horizontal and vertical components of the stresses in the members c a and ch. Bin the member ce is horizontal, it will have no vertical component. The vertical and horizontal components of all the stresses in the oblique members may be obtained by scaling the lines which represent them. Since all the members except the right-hand oblique one are in pairs, one-half of each stress will be carried on each side of the center line.

Having obtained the amounts of the horizontal and vertical components, the diagrams (c) and (d) may be drawn. Care must be exercised to see that all the forces upon one side of the pin are equal to those upon the other; otherwise, the joint would not be in equilibrium, and would tend to move in the direction of the greater force. The diagram (c) shows the horizontal components of all the forces, the bending moment due to which is obtained as follows: (For bending moment see page 111.)

22,050 lb. X 5 in. = 110,250 in.-lb.

deduct, 3,000 lb. X 3 in. = 9,000 in.-lb.

15,000 lb. X 4 in. = 60,000 in.-lb. 69,000 in.-lb.

Total horizontal bending moment = 41,250 in.-lb. From the diagram (d) of all the vertical components the bending moment due to them is obtained thus:

1,800 lb. X 5 in. = 9,000 in.-lb.

5,150 lb. X 3 in. = 15,450 in.-lb.

Total vertical bending moment = 24,450 in.-lb. The resultant bending moment is, then:

Strength Of Pins 271 = 47,900 in.-lb.

From Table XXX, page 137, using a safe unit stress of 15,000 lb., it will be seen that a pin which is 3 3/16 in. in diameter has a resisting moment of 47,670 in-lb., which, while scant, will do.

If the pin is designed to resist bending, it is seldom neces-aary to consider the shear. The bearing values are found the same as for rivets, and are given in Table XXX.