Shear and Tensile force with factor of safety

[chris78]

1. Homework Statement
A load p of 5kn is applied to the tensile member shown and carried at the joint by a single 20mm diameter rivet.The angle of the shear joint is 60 degrees to the axis of the load.

calculate the tensile stress in the rivet
calculate the shear stress in the rivet

given that the ultimate tensile strength for the rivet is 80MN/m. What is the safety factor for this joint.

Homework Equations

The Attempt at a Solution

shear force=5sin30
shear force=2.5kn

Tensile force=5cos30
Tensile force=4.330127knCSA of rivet=pie x 0.01^2
CSA of rivet=3.14159 x 10^-4

Shear stress=force/area

Shear stress=2500N/3.14159 x 10^-4
Shear stress=7957753.876 NM/2

Tensile stress=4330.127N/3.14159 x 10^-4
Tensile stress=13783233.97 NM/2

Factor of safety= UTS/working stress

80MN/m=80 x 10^6 NM/2

Shear=80x10^6/7957753.976=10.053
Tensile=80x10^6/13783233.97=5.804

Factor of safety for joint=5.8

I posted a similar question yesterday and thank all for there help...i just wanted to make sure that the above is correct and I am resolving my vectors correctly to work out the shear and tensile forces.

Thanks again

Chris

 

[PhanthomJay]

Your numbers for shear and tensile stress look good, but be careful on how you write your units. A megaNewton per meter squared stress is MN/m², usually denoted as MPa ((megapascal) for short. Don't express large numbers in Newtons, they get too long. Also be sure to round off all answers to one decimal point at the most. Your safety factor for tension stress also looks good. Don't indicate any safety factor for shear stress, as not only it was not asked,but also the ultimate shear strength for the bolt is a lot less than the ultimate tensile strength.

 

[chris78]

Your numbers for shear and tensile stress look good, but be careful on how you write your units. A megaNewton per meter squared stress is MN/m², usually denoted as MPa ((megapascal) for short. Don't express large numbers in Newtons, they get too long. Also be sure to round off all answers to one decimal point at the most. Your safety factor for tension stress also looks good. Don't indicate any safety factor for shear stress, as not only it was not asked,but also the ultimate shear strength for the bolt is a lot less than the ultimate tensile strength.

Hi there

Thanks again for your input...i will watch my units and decimal places...I have attached a copy of my working...and specifically my triangle vector...can you confirm that I am setting this up correctly,as when i handed a similar question in he said my triangle was wrong,and therefore my shear and tensile force.

Thanks for your help

 

[PhanthomJay]

Hi there

Thanks again for your input...i will watch my units and decimal places...I have attached a copy of my working...and specifically my triangle vector...can you confirm that I am setting this up correctly,as when i handed a similar question in he said my triangle was wrong,and therefore my shear and tensile force

Thanks for your help

Your diagram is wrong, even though your answer is correct. Always sketch your diagrams close to scale. When breaking up a vector into its components, use the arrow to tail graphical sketch as shown. Now I understand why your solution is being marked wrong. See attached

 

[PhanthomJay]

Actually, your sketch will be ok if you show the bolt, change the vector arrows accordingly, and show the angles to a better scale (it looks like 45 degree angles instead of 30 and 60). Otherwise your answers for tension and shear are ok.