Question

Hey all, I just need help for the start of this question!

The assembly is used to support the distributed loading of w=10 kN/m. Determine the factor of safety with respect to yielding for the steel rod BC and the pins at B and C if the yield stress for the steel in tension is σy=250 MPa and in shear τy=125 MPa. The rod has a diameter of 13 mm, and the pins each have a diameter of 10 mm.

I'm having trouble finding out P and A, as finding the factor of safety is simple enough.

Answer 1

Here's an attachment of the picture...

Answer 2

@YanaSidlinskiy

Answer 3

@Abhisar Any ideas? Yana recommended to tag you in it in hopes that you're able to help me out.

Answer 4

But you're currently offline :p

Answer 5

@Miracrown

Answer 6

Everyone's offline :p Just my luck...

Answer 7

I've looked over lecture materials as well as a US version of this question (using inches) but I can not for the life of me figure out where they got their numbers from...

Answer 8

@Mathmale

Answer 9

@Kainui ? :(

Answer 10

Is it above A level physics?

Answer 11

I had it as college physics, but I feel like the problem I'm having with the question is around high school standard... I just couldn't for the life of me figure out how the internet got P..

Answer 12

http://www.slideshare.net/alvarovelazquezThe581/ch01-03-stress-strain-properties
Solution if you're interested (except mine is not in inches) Problem number 1-93 so you have to scroll to the bottom for it.

Answer 13

@ganeshie8 Any ideas?

Answer 14

start by taking moments about A

so \((0.9)T \sin (A \hat B C) = 0.6W\) where T is tension in rod, W is the weight of the beam = \(10kN/m \times 1.2m = 1200N\)

solve for T and \(\sigma_{actual} = \dfrac{0.6 \times 1200}{(0.9)\sin (A \hat B C) \times A} \) where A is cross sect area of rod

it's a 3-4-5 pythagorean triple so \(\sin (A \hat B C) = 4/5\) etc

you might even get this as your answer ..... or might not!

https://www.wolframalpha.com/input/?i=%3D+250000000+%2F+%7B+(0.6*1200)+%2F+(0.9*(1.2%2F1.5)+*+(pi%2F4)+*+0.013%5E2)%7D



NB the shearing bit involves just a wee bit more work

get the vertical shear force by this time taking moments about B, using same procedure as above

the horizontal component of shear is just \(T \cos \theta\). combine vert and horiz components using Pythagoreas, then divide by cross sect area of the pin to get actual shear stress. there seems to be just one pin.

Answer 15

I understand it now, THANK YOU SO MUCH! :)