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Stress 4

The document discusses the concept of shear stresses in materials, explaining how they arise from parallel forces and induce complementary shear stresses for equilibrium. It includes examples of calculating maximum average bearing stress and average shear stress in bolted connections and structural members. Additionally, it introduces the factor of safety, emphasizing the importance of designing structural components to withstand stresses below their ultimate strength.

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yw89407
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37 views12 pages

Stress 4

The document discusses the concept of shear stresses in materials, explaining how they arise from parallel forces and induce complementary shear stresses for equilibrium. It includes examples of calculating maximum average bearing stress and average shear stress in bolted connections and structural members. Additionally, it introduces the factor of safety, emphasizing the importance of designing structural components to withstand stresses below their ultimate strength.

Uploaded by

yw89407
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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CHAPTER 1

CONCEPT OF STRESS
1.4 Shear Stresses:

Where the cross – sectional area


of a block of material is subject
to a distribution of forces which
are parallel.
The resulting force intensities are
known as shear stresses

𝑷
𝝉=
𝑨
The existence of shear stresses on any two sides of the
element induces complementary shear stresses on the
other two sides of the element to maintain equilibrium.
As shown in the figure the shear stress τ in sides AB and
CD induces a Complimentary shear stress τ’ in sides AD
and BC.
Sign convections for shear stresses:
- Tending to turn the element C.W +𝑣𝑒.
- Tending to turn the element C.C.W −𝑣𝑒.
Example 1.2
Consider the double shear bolted connection at right. The upper and lower
plates have thickness of a = 10 mm. The middle plate has thickness b = 15
mm. The bolt has diameter d = 10 mm. The system is subjected to a load of
P = 5 KN as shown at right.

Required:
a) The Maximum Average Bearing Stress in the bolt.
b) The Average Shear Stress in the bolt.

Solution:
the Average Bearing Stresses on the top and bottom
surfaces of the bolt are:
The Average Bearing Stresses on the middle surface of
the bolt:

Therefor, the maximum average Bearing Stresses is 𝜎𝐵,𝑚𝑎𝑥. = 33 𝑀𝑃𝑎

The average shear stress occur in the top or bottom of the bolt
𝑃/2 2.5×103
𝜏𝑠 = = = 32 𝑀𝑃𝑎
𝜋𝑟 2 𝜋52
Example 1.3
1.5 Factor of Safety (FS)
Structural members or machines must be designed such that the
working stresses are less than the ultimate strength of the
𝜎𝑢 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
material FS = =
𝜎𝑎𝑙𝑙 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
Example 1.4
Example 1.5
Determine the maximum vertical force P that can be applied
to the bell crank so that the average normal stress
developed in the 10-mm diameter rod CD, and the average
shear stress developed in the 6-mm diameter double
sheared pin B not exceed 175 MPa and 75 MPa, respectively.
𝜋 𝜋 2
𝐴𝑐𝑑 = 102 = 78.54 𝑚𝑚 2
, 𝐴𝐵 = 6 = 28.27 𝑚𝑚2
4 4

For link CD the normal stress


𝐹 𝐹𝑐𝑑
𝜎𝑐𝑑 = 𝑐𝑑 , 175 = , 𝐹𝑐𝑑 = 13.75 𝐾𝑁
𝐴𝑐𝑑 78.54

For equilibrium of the link ABC

σ 𝑀𝐵 = 0 , 𝐹𝐶𝐷 300𝑠𝑖𝑛45 − 450𝑃 = 0 ,

13.75 300𝑠𝑖𝑛45 − 450𝑃 = 0 , 𝑃 = 6.48 KN

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