Calculating Safe Washer Diameter for 20mm Bolt in Strength of Materials Seatwork

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SUMMARY

The calculation of the safe washer diameter for a 20 mm bolt under a tensile stress of 150 MPa reveals discrepancies between student and teacher results. The teacher's calculated diameter of 70.80 mm is incorrect due to multiple errors, including misreading values and using incorrect cross-sectional areas. The correct formula to determine the washer diameter is d = [(σ/Sb)D² + di²]⁰.⁵, leading to a safe diameter of 63.88 mm when using the appropriate parameters. Key factors include the correct tensile stress area for the bolt and the typical inside diameter of the washer.

PREREQUISITES
  • Understanding of tensile stress and bearing stress concepts
  • Familiarity with the geometry of washers and bolts
  • Knowledge of the formula for cross-sectional area of circular shapes
  • Proficiency in unit symbols and their correct capitalization
NEXT STEPS
  • Study the calculation of tensile stress areas for various bolt sizes
  • Learn about the geometry and specifications of flat washers
  • Research common mistakes in engineering calculations and how to avoid them
  • Explore the implications of incorrect unit usage in engineering contexts
USEFUL FOR

Mechanical engineers, students in strength of materials courses, and professionals involved in fastening and structural integrity assessments will benefit from this discussion.

dtenyente
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Seatwork no. 1 -------Strength of Materials----------

The 20 mm diameter bolt fastens two wooden planks together. The nut is tightened until the tensile stress in the bolt is 150 Mpa. Find the smallest safe diameter “d” of the washers, if the working bearing stress for wood is 13 Mpa.

Note: My teacher said that the
“d” = 70.80mm


My solution:
Pb = Ab Sb
P = 150 N/mm2 * ( π/14 * 20 mm2 )
P = 47,123.89 N

47,123.89 N = π/14 (d2) 13
d = 67.94
Question: My “d” is not equal to the “d” of my teacher. Pls help me.
 

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The equation "P = 150 N/mm2 * ( π/14 * 20 mm2 ) " is a bit awkward.

For a circular cross-section, A = π r2 = π/4 d2, so the π/14 doesn't seem to make sense.

The load P = A * S, where A is the cross-sectional area and S is the stress.

Basically the tensile load in the bolt (shank) must equal the compressive load on the wood. But the two surface areas will be different.

So what is the geometry of a washer, and to what does diameter of a washer refer?
 
Well that's my main problem, maybe lack of information in the problem. I'm not sure with my solution. All of my classmates didn't get the right answer as what my teacher said. Maybe my teacher forgot to write some details in the drawing. Am i right?
 
dtenyente: You made a few mistakes. Here's what you computed (basically).

d = [(150/13)20^2 + 0^2]^0.5 = 67.94 mm.

Your teacher made five mistakes. Here's how he computed his answer, followed by an explanation of each mistake.

d = [(150/13.006)20^2 + 20^2]^0.5 = 70.80 mm.

(1) He divided 150 by 13.006. It appears he misread his calculator (which said 11.538), and thought it said 11.533.

(2) He used the wrong cross-sectional area of the bolt. If someone says, "The bolt tensile stress is 150 MPa," it means the tensile stress is 150 MPa on the bolt tensile stress area, not the shank. The tensile stress diameter for an M20 bolt is D = 17.655 mm, not 20 mm.

(3) He assumed the washer inside diameter is equal to the bolt nominal diameter, whereas the inside diameter of an M20 flat washer is occasionally 21 mm, but usually 22 mm, or sometimes more.

(4) He failed to state the washer inside diameter in the problem statement. Therefore, the best assumption is to use the typical inside diameter of an M20 flat washer, which appears to be 22 mm. Let's assume di = 22 mm.

(5) He misspelled MPa, or maybe that was you. You must use the correct capitalization for each letter in a unit symbol; otherwise, it generally means something else, which is wrong.


Here is the correct answer.

d = [(sigma/Sbr)D^2 + di^2]^0.5
= [(150/13)17.655^2 + 22^2]^0.5 = 63.88 mm.
 

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