BEARING STRESS

BARRIENTOS, Lei AnneMARTIREZ, WilburMORIONES, Jan EbenezerNERI, Laiza Paulene

MAPUA INSTITUTE OF TECHNOLOGY

MEC32/A1

Members:

TOPICS

THEORY ON BEARING STRESS

FORMULA FOR BEARING STRESS

EXAMPLES ON BEARING STRESS

Bearing stress is a contact pressure between separate bodies. It differs from compressive stress because compressive stress is the internal stress caused by a compressive force.

FORMULA:

𝜎 𝑏=𝑃𝑏

𝐴𝑏Where:

𝑃𝑏−𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒𝑙𝑜𝑎𝑑

𝐴𝑏− h𝑐 𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑎𝑟𝑒𝑎𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑃𝑏

𝜎 𝑏−𝑏𝑒𝑎𝑟𝑖𝑛𝑔𝑠𝑡𝑟𝑒𝑠𝑠

DIFFERENCE BETWEEN NORMAL, SHEAR AND BEARING STRESSES

Normal Stress – stress normal to the surface

Shearing Stress – stress tangent to the surface

Bearing Stress – compressive force divided by the characteristic area perpendicular to it

Bearing Stress – compressive force divided by the characteristic area perpendicular to it

In Fig. 1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; and (b) the largest average tensile stress in the plates.

PROBLEM 1

a) From shearing of rivet:

𝑃=𝜏 𝐴𝑟𝑖𝑣𝑒𝑡𝑠=60 [ 14 𝜋 (20)2]=6000𝜋𝑁 From bearing of plate material:

𝑃=𝜎𝑏 𝐴𝑏

𝑡=7.85𝑚𝑚

b) Largest average tensile stress in the plate: 𝑃=𝜎 𝐴

6000𝜋=𝜎 [7.85(110−20) ]

SOLUTION

In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.

PROBLEM 2

For shearing of rivets (double shear):

14=20 [2(0.8618 𝑡) ]

14=12 [2 ( 14 𝜋 𝑑2)]

For bearing of yoke:𝑃=𝜎𝑏 𝐴𝑏

SOLUTION

The lap joint shown in the figure is fastened by three 20 mm. diameter rivets. Assuming that P=50 Kn.1. Determine the shearing stress in each rivet.

2. Determine the bearing stress in each plate.

3. Determine the maximum average tensile stress in each plate. Assume that the axial load P is distributed equally among the three rivets.

PROBLEM 3

SOLUTION

1. Shearing stress in each rivet

2. Bearing Stress in each plate

3. Maximum tensile stress in each plate

𝑆𝑚𝑎𝑥=𝑃𝐴𝑛𝑒𝑡

𝐴𝑛𝑒𝑡=(130−20 ) (25 )=2750𝑚𝑚2

For the lap joint shown in the figure.

1. Determine the maximum safe load P which may be applied if the shearing stress in the rivets is limited to 60 MPa.

2. Determine the safe load P which may be applied if the bearing stress of the plate is limited to 110 Mpa.3. Determine the safe load P if the average tensile stress of the plate is limited to 140 Mpa.

PROBLEM 4

1. Safe load P due to shear of rivets

2. Load P due to bearing of plates

3. Load P due to tearing of plates

P =

𝐴𝑛𝑒𝑡=(130−20 ) (25 )=2750𝑚𝑚2

Therefore, maximum safe load, P=56 5 N (shearing of rivets govern).

SUMMARY

Bearing stress is a contact pressure between separate bodies.

Where:

 

 

 

REFERENCES

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand L. Singer

Mechanics of Materials 2nd Edition by Andrew Pytel and Jaan Kiusalaas“How materials carry load?” http://emweb.unl.edu/Department of Engineering Mechanics, University of Nebraska, Lincoln, NELast modified at: 3:07 PM, Wednesday, August 30, 2000