TORSION FLANGED BOLT COUPLINGS

In shaft connection called flanged bolt couplings, the torque is transmitted by the shearing force P created in the

bolts that is assumed to be uniformly distributed. For any number of bolts n, the torque capacity of the coupling is:

If a coupling has two concentric rows of bolts, the torque

capacity is,

where the subscript 1 refers to bolts on the outer circle and

subscript 2 refers to bolts on the inner circle.

For rigid flanges, the shear deformations in the bolts are

proportional to their radial distances from the shaft axis. The

shearing strains are related by,

Using Hooke’s law for shear, 𝐺 = 𝜏/𝛾, we have

If the bolts on the two circles have the same area, 𝐴1 = 𝐴2, and if the bolts are made of the same material, 𝐺1 = 𝐺2, the

relation between 𝑃1 𝑎𝑛𝑑 𝑃2 reduces to:

A flanged bolt coupling consists of ten 20 mm diameter bolts

spaced evenly around a bolt circle 400 mm in diameter.

Determine the torque capacity of the coupling if the allowable shearing stress in the bolt is 40 MPa.

PROBLEM 1

PROBLEM 1 SOLUTION

A flanged bolt coupling consists of eight 10 mm diameter

steel bolts on a bolt circle 400 mm in diameter, and six 10 mm

diameter steel bolts on a concentric bolt circle 300 mm in

diameter, as shown below. What torque can be applied without exceeding a shearing stress of 60 MPa in the bolts?

PROBLEM 2

PROBLEM 2 SOLUTION

In a rivet group subjected to a twisting couple T, show that

the torsion formula 𝜏 = 𝑇𝜌/𝐽 can be used to find the shearing

stress 𝜏 at the center of any rivet. Let 𝐽 = 𝐴𝜌2, where 𝐴 is the

area of a rivet at the radial distance 𝜌 from the centroid of

the rivet group.

PROBLEM 3

PROBLEM 3 SOLUTION

Six 7/8 in diameter rivets fasten the plate to the fixed

member as shown below. Using the result in Problem 3,

determine the average shearing stress caused in each rivet by the 14 kip loads. What additional loads P can be applied

before the shearing stress in any rivet exceeds 8,000 psi?

PROBLEM 4

PROBLEM 4 SOLUTION

PROBLEM 4 SOLUTION

PROBLEM 4 SOLUTION

TORSION THIN-WALLED TUBE

The torque applied to thin-walled tubes is expressed as,

The average shearing stress across any thickness t is,

Thus, torque T can also be expressed as,

A torque of 600 N-m is applied to the rectangular section

shown below. Determine the wall thickness t so as not to

exceed a shear stress of 80 MPa.

PROBLEM 5

PROBLEM 5 SOLUTION

A tube 2 mm thick has the shape shown below. Find the

shearing stress caused by a torque of 600 N.m.

PROBLEM 6

PROBLEM 6 SOLUTION

TORSION HELICAL SPRINGS

When close-coiled helical spring, composed of a wire of round rod of diameter d wound into a helix of mean radius R

with n number of turns, is subjected to an axial load P

produces the following stresses and elongation:

The maximum shearing stress is the sum of the direct

shearing stress 𝜏1 = 𝑃/𝐴 and the torsional shearing stress

𝜏2 = 𝑇𝑟/𝐽, with 𝑇 = 𝑃𝑅.

This formula neglects the curvature of the spring. This is used

for light spring where the ratio 𝑑/4𝑅 is small.

For heavy springs and considering the curvature of the spring, a more precise formula is given by (A.M. Wahl formula):

The elongation of the bar is,

Notice that the deformation 𝛿 is directly proportional to the applied load P. The ratio of P to 𝛿 is called the spring constant

𝑘 and is equal to,

For two or more springs with springs laid in series, the resulting

spring constant 𝑘 is given by,

For two or more springs in parallel, the resulting spring

constant 𝑘 is given by,

Determine the maximum shearing stress and elongation in a helical steel spring composed of 20 turns of 20 mm diameter

wire on a mean radius of 90 mm when the spring is

supporting a load of 1.5 kN. Use A.M. Wahl formula, and G =

83 MPa.

PROBLEM 7

PROBLEM 7 SOLUTION

Two steel springs arranged in series as shown supports a load P. The upper spring has 12

turns of 25 mm diameter wire on a mean radius

of 100 mm. The lower spring consists of 10 turns

of 20 mm diameter wire on a mean radius of

75 mm. If the maximum shearing stress in either

spring must not exceed 200 MPa, compute

the maximum value of P and the total

elongation of the assembly. Use A.M. Wahl

formula, and G = 83 GPa. Compute the

equivalent spring constant by dividing the

load by the total elongation.

PROBLEM 8

PROBLEM 8 SOLUTION

PROBLEM 8 SOLUTION

As shown below, a homogenous 50-kg rigid block is

suspended by three springs whose lower ends were originally at the same level. Each steel spring has 24 turns of 10 mm

diameter on a mean diameter of 100 mm, and G = 83 GPa.

The bronze spring has 48 turns of 20 mm diameter wire on a

mean diameter of 150 mm, and G = 42 GPa. Compute the

maximum shearing stress in each spring.

PROBLEM 9

PROBLEM 9 SOLUTION

PROBLEM 9 SOLUTION

PROBLEM 9 SOLUTION

PROBLEM 9 SOLUTION