Factor of Safety - Mechanics of Materials

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Discussion Overview

The discussion revolves around a homework problem related to the mechanics of materials, specifically focusing on determining the minimum diameters required for various components of a rigid beam subjected to a uniformly distributed load. Participants are analyzing the factor of safety, shear strength, and yield strength of materials involved in the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the resultant force of the distributed load as 14.3 kN/m and attempts to find the x and y components of the force on the tie rod.
  • Another participant questions the correctness of the calculations, suggesting that the resultant load acts at the center of gravity when determining moments.
  • Concerns are raised about the use of incorrect lever arms and the handling of forces and moments in the calculations.
  • There is a suggestion that the force at point C should be calculated using vector sums rather than algebraic sums.
  • Multiple participants express uncertainty about the correctness of the calculated diameters for the tie rod and pins, indicating that the values may be incorrect and need recalculation.
  • One participant acknowledges a mistake in typing the lever arm length and discusses the potential errors in calculating stress in the pins.
  • Another participant emphasizes the importance of correctly applying the factor of safety and shear strength in the calculations.

Areas of Agreement / Disagreement

Participants generally agree that there are errors in the calculations, but specific disagreements remain regarding the correct approach to determining the forces and moments. The discussion is unresolved, with multiple competing views on the correct methodology and calculations.

Contextual Notes

Participants note potential limitations in their calculations, including assumptions about the application of forces, the handling of double shear conditions, and the accuracy of lever arm lengths. There is also mention of the need for careful attention to mathematical signs and components in the calculations.

Triathlete
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Homework Statement



Rigid beam ABC is supported as shown in the figure. The pin connections at B, C, and D are each double shear connections, and the ultimate shear strength of the pin material is 640 MPa. Tie rod (1) has a yield strength of 350 MPa. A uniformly distributed load of w = 11 kN/m is applied to the beam as shown. A factor of safety of 3.5 is required for all components. Assume a = 575 mm, b = 725 mm, c = 250 mm, and d = 525 mm.

Determine

(a) the minimum diameter required for tie rod (1).
(b) the minimum diameter required for the double shear pins at B and D.
(c) the minimum diameter required for the double shear pin at C.

http://edugen.wileyplus.com/edugen/courses/crs6838/art/qb/qu/ch0/13425408508040_6521478752676759.jpg

Homework Equations



σ = F/A
τ = F/A
σallow = σfailure/FS
τallow = τfailure/FS

The Attempt at a Solution



First I got the resultant force of w → 14.3 kN/m.
I started by finding the x and y components of F1(force on the rod), which were F1cos35.9 and F1sin35.9, respectively.
Then I summed the moments about C which gave me ΣM = (-14.3 kN/m)(13m) + (0.810F1)(0.25m) = (0.5864F1)(0.725)
⇒ F1 = 14.8 kN
Using σ ≤ σallow, I plugged in my values → 14809.4 N/π(d/2)2 ≤ 350 N/mm2/3.5
⇒ drod = 188.6 mm

14809.4 N/2π(d/2)2 ≤ 640 N/mm2/3.5
⇒dB,D = 51.6 mm

Then I summed the x and y forces to get the force at C, which was C = 13.2 kN

13242.3 N/2π(d/2)2 ≤ 640 N/mm2/3.5
⇒ d = 46.1 mm

All of my d values were wrong. I have recalculated and changed the signs on my forces and moments just to see if I did that wrong but I still can't get the right answer. Am I going about the problem the wrong way? I'm just not sure where I'm going wrong and I've been working on this and a similar problem for a couple days now to no avail.
 
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Triathlete said:

Homework Statement



Rigid beam ABC is supported as shown in the figure. The pin connections at B, C, and D are each double shear connections, and the ultimate shear strength of the pin material is 640 MPa. Tie rod (1) has a yield strength of 350 MPa. A uniformly distributed load of w = 11 kN/m is applied to the beam as shown. A factor of safety of 3.5 is required for all components. Assume a = 575 mm, b = 725 mm, c = 250 mm, and d = 525 mm.

Determine

(a) the minimum diameter required for tie rod (1).
(b) the minimum diameter required for the double shear pins at B and D.
(c) the minimum diameter required for the double shear pin at C.

http://edugen.wileyplus.com/edugen/courses/crs6838/art/qb/qu/ch0/13425408508040_6521478752676759.jpg

Homework Equations



σ = F/A
τ = F/A
σallow = σfailure/FS
τallow = τfailure/FS


The Attempt at a Solution



First I got the resultant force of w → 14.3 kN/m.
You mean 14.3 kN
I started by finding the x and y components of F1(force on the rod), which were F1cos35.9 and F1sin35.9, respectively.
Then I summed the moments about C which gave me ΣM = (-14.3 kN/m)(13m) + (0.810F1)(0.25m) = (0.5864F1)(0.725)
⇒ F1 = 14.8 kN
Using σ ≤ σallow, I plugged in my values → 14809.4 N/π(d/2)2 ≤ 350 N/mm2/3.5
⇒ drod = 188.6 mm

14809.4 N/2π(d/2)2 ≤ 640 N/mm2/3.5
⇒dB,D = 51.6 mm

Then I summed the x and y forces to get the force at C, which was C = 13.2 kN

13242.3 N/2π(d/2)2 ≤ 640 N/mm2/3.5
⇒ d = 46.1 mm

All of my d values were wrong. I have recalculated and changed the signs on my forces and moments just to see if I did that wrong but I still can't get the right answer. Am I going about the problem the wrong way? I'm just not sure where I'm going wrong and I've been working on this and a similar problem for a couple days now to no avail.
You are getting a bit careless I think on math and signage, and using the wrong lever arm for the distributed load moment. The beam is 1.3 m long. Its resultant load acts at its cg when determining moments. The pins are in double shear.
 
PhanthomJay said:
You mean 14.3 kNYou are getting a bit careless I think on math and signage, and using the wrong lever arm for the distributed load moment. The beam is 1.3 m long. Its resultant load acts at its cg when determining moments. The pins are in double shear.

Oops I meant .65m not 1.3m. I calculated it using the 0.65 but I typed it wrong. I also thought I was calculating the stress in the pins correctly, by dividing the force by 2. Where else am I going wrong?
 
I will have to check your numbers I haven't done that ...when you say your force at C is the sum of the x and y components I hope you meant vector sum not algebraic sum. You have the 2 in there for dbl shear so that's ok , your approach is good I'll check math later
 
Your math is way off when calculating the diameters. Try again.
 

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