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

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Discussion Overview

The discussion revolves around calculating the safe diameter of washers for a 20 mm bolt used to fasten wooden planks, considering tensile stress in the bolt and bearing stress for wood. Participants explore the mathematical approach and the assumptions involved in the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The original poster presents a calculation for the washer diameter based on given tensile stress and bearing stress, arriving at a different value than their teacher.
  • One participant questions the formulation of the original equation, suggesting that the area calculation for the bolt's cross-section may be incorrect.
  • Another participant expresses uncertainty about the problem's completeness, suggesting that missing details may have led to discrepancies in the answers.
  • A later reply identifies several mistakes in the teacher's calculation and provides an alternative formula for determining the washer diameter, citing specific values for the bolt's tensile stress area and washer dimensions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct diameter for the washers, with multiple competing views and calculations presented. The discussion remains unresolved regarding the correct approach and values to use.

Contextual Notes

Limitations include potential misinterpretation of the problem statement, assumptions about washer dimensions, and the need for clarity on the cross-sectional area used in calculations.

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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