- #1

gpsza

- 2

- 0

Below is a self assesment question I found in a study guide related direct and shear stress, it is in 2 parts. The guide only gives the answers. I managed part 1 but for part 2 my answer differs from the given safety factor solution of 7 mine is 12.5. In my attempt below i have tried to give a description of the approach I used.

I would appreciate if someone could point out the error in my approach.

Thanks in advance

gpsza

## Homework Statement

We have 2 steel rods connected using a clevis joint.

Part 1. Determine maximum allowed force using given shear stress safety factor

Part 2. Based on answer of part 1 determine the safety factor for the rods

variables:

pin diameter = 8 mm

ultimate shear stress of pin material = 80 Mpa or Alternatively 80 N/(mm*mm)

shear stress Safety Factor = 2

rod material yield stress = 160 Mpa or Alternatively 160 N/(mm*mm)

rod diameter = 20 mm

## Homework Equations

Part 1:

(working shear stress)=(Maximum Shear Stress)/(Safety Factor)

Force = (shear stress)*(2*(Cross sectional area of pin))

(Circle Area)=(pi*diameter*diameter)/4

Part 2:

(direct stress)=Force/Area

(safety factor)=(Stress at Failure)/(Maximum Working Stress)

## The Attempt at a Solution

Part 1:

Calculate working stress:

(working shear stress) = 80/2 = 40 Mpa or Alternatively 40 N/(mm*mm)

Calculate force keeping in mind the need to double the area of the pin:

Area = (pi*8*8)/4 = 50.2654824574 (mm*mm)

2*Area = 100.5309649149 (mm*mm)

Force = 40 * 100.5309649149 = 4021.2385965949 = 4.02 kN

Part 2:

Approach: use force calculated in part 1 to calculate the direct stress the rods would experience, use the direct stress and given yield stress to calculate the safety factor

Area = (pi*20*20)/4 = 314.159265359 (mm*mm)

(direct stress) = 4021.2385965949 / 314.159265359 = 12.8 Mpa

(safety factor)=(Stress at Failure)/(Maximum Working Stress)

(safety factor)= 160/12.8 = 12.5