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# Bearing stress

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This powerpoint presentation deals mainly about bearing stress, its concept and its applications.

Members:
BARIENTOS, Lei Anne
MARTIREZ, Wilbur
MORIONES, Jan Ebenezer
NERI, Laiza Paulene

Sir Romeo Alastre - MEC32/A1

Published in: Engineering
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• Slide 8 bearing of the plate should be 6000pi= 120*(t*20)

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• this was so helpful :)

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• You just took into account the thickness not the whole are of the normal surface in slide 8. You should put your units so it is easier for people to follow along and so that you don't make errors like that.

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### Bearing stress

1. 1. BARRIENTOS, Lei Anne MARTIREZ, Wilbur MORIONES, Jan Ebenezer NERI, Laiza Paulene MAPUA INSTITUTE OF TECHNOLOGY MEC32/A1 Members:
2. 2. TOPICS  THEORY ON BEARING STRESS  FORMULA FOR BEARING STRESS  EXAMPLES ON BEARING STRESS
3. 3. 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.
4. 4. FORMULA: 𝜎 𝑏 = 𝑃𝑏 𝐴 𝑏 Where: 𝑃𝑏 − 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 𝐴 𝑏 − 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑎𝑟𝑒𝑎 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜𝑃𝑏 𝜎𝑏 − 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
5. 5. 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
6. 6. Bearing Stress – compressive force divided by the characteristic area perpendicular to it
7. 7. 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
8. 8. a) From shearing of rivet: 𝑃 = 𝜏𝐴 𝑟𝑖𝑣𝑒𝑡𝑠 = 60 1 4 𝜋(20)2 = 6000𝜋 𝑁 From bearing of plate material: 𝑃 = 𝜎𝑏 𝐴 𝑏 𝑡 = 7.85 𝑚𝑚 b) Largest average tensile stress in the plate: 𝑃 = 𝜎𝐴 6000𝜋 = 𝜎 7.85(110 − 20) SOLUTION
9. 9. 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
10. 10. For shearing of rivets (double shear): 𝑃 = 𝜏𝐴 14 = 20 2(0.8618𝑡) 14 = 12 2 1 4 𝜋𝑑2 For bearing of yoke: 𝑃 = 𝜎𝑏 𝐴 𝑏 SOLUTION
11. 11. 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
12. 12. SOLUTION 1. Shearing stress in each rivet 2. Bearing Stress in each plate 3. Maximum tensile stress in each plate 𝑆 𝑚𝑎𝑥 = 𝑃 𝐴 𝑛𝑒𝑡 𝐴 𝑛𝑒𝑡 = 130 − 20 25 = 2750 𝑚𝑚2
13. 13. 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
14. 14. 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 = 𝐴 𝑛𝑒𝑡 𝑆 𝐴 𝑛𝑒𝑡 = 130 − 20 25 = 2750 𝑚𝑚2 Therefore, maximum safe load, P=56 5 N (shearing of rivets govern).
15. 15. SUMMARY  Bearing stress is a contact pressure between separate bodies. Where:
16. 16. 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, NE Last modified at: 3:07 PM, Wednesday, August 30, 2000