# Factor of safety

1. Nov 23, 2007

### series111

1. The problem statement, all variables and given/known data
TWO ALUMINIUM STRIPS EACH 75mmx5mm ARE RIVETED TOGETHER TO FORM A CONTINUOUS LENGTH THE JOINT IS MADE USING 6 RIVETS EACH 5mm IN DIA THE ULTIMATE SHEAR STRENGTH OF THE RIVET MATERIAL IS 70MN/m2 AND THE UTS OF THE ALUMINIUM IS 30MN/m2 IF THE MAXIMUM TENSILE FORCE APPLIED TO THE CONTINUOUS STRIP IS 1.875KN DETERMINE THE FACTOR OF SAFETY FOR (a) THE RIVETS (b) THE ALUMINIUM

2. Relevant equations

3. The attempt at a solution
area for rivets piex5x10-32/4 x 6 =1.178x10-4 area for aluminium 75x10-3x5x10-3 x 2 = 750 x 10-6 shear stress for rivets = 1.875x10 3/1.178x10-4=15.91x10 6 shear stress for aluminium=1.875x10 3 / 750 x 10-6=2.5x 10 6 factor of safety for aluminium=30x10 6 / 2.5 x 10 6 = 12 factor of safety for rivets = 70 x 10 6/ 15.91 x 10 6 = 4.39

2. Nov 23, 2007

### PhanthomJay

I didn't check your math, but I note that you are looking for the tensile stress in the aluminum, not shear as you indicate, , and that you have used the wrong area in determining that tensile stress (you're off by a multiplicative factor).

3. Nov 23, 2007

### series111

where have i went wrong as i thought this was correct?

4. Nov 23, 2007

### PhanthomJay

Your error is noted in red

5. Nov 24, 2007

### series111

thanks i did only one plane first but then i multiplyed it by 2 because i got confused with shear stress thanks again i will post my answers and if you would be so kind to check them i will be most gratefull as this is part of my first mechanical principles assessment.

6. Nov 24, 2007

### series111

area of rivets = pie x 5x10-3 2/4 x 6 =1.178 x 10 -4

area of aluminium= 75x 10 -3 x 5 x 10 -3 = 0.375 x 10 -3

tensile stress of rivets =1.875 x 10 3 / 1.178x 10-4 = 15.91 x 10 6

tensile stress of aluminium = 1.875 x 10 3 / 0.375 x 10 -3 = 5 x 10 6

factor of safety for rivets = 70 x 10 6 / 15.91 x 10 6 = 4.39

factor of safety for aluminium = 30 x 10 6 / 5 x 10 6 = 6

7. Nov 24, 2007

### PhanthomJay

Your numbers are now correct, but you are confusing the differences between shear and tensile stresses. The rivets are in shear, the aluminum strips are in tension. Picture two strips of aluminum, each say 1 meter long, and riveted together at one overlapping end with 6 rivets, so that now the strip is almost 2 meters long (it is slightly less than 2 meters due to the overlap, but the length does not matter in this problem, I just want to be sure you see the problem correctly). Now you apply the 1.875kN tensile load, which puts the strips in tension (force perpendicular to the cross sectional area of the strip causing tensile stress), and puts the rivets in shear (force parallel to the area of the rivets, causing shear stress).

8. Nov 24, 2007

### series111

thanks very much i am trying to cram so much information in as i have just returned to education after 10 year spell off and i am confusing myself that why i joined this forum as i need all the help i can get thanks again mark.

9. Jan 21, 2008

### series111

hi again just got my assessment back and all figures are correct but the factor safety formula for the rivets is wrong this is what a put fs= uts/stress and i dont no whyit is wrong can you help as it is confusing me hope you can help again thanks mark.

10. Jan 21, 2008

### PhanthomJay

I don't know either unless it's due to not rounding off; FS in rivets is 4.398 perhaps they are looking for 4.4 or even just 4? Your numbers and method look correct .

11. Jan 21, 2008

### series111

The answer is correct it is just the uts that wrong because in the question it states the ultimate shear strength of the rivet material is 70MN/m2 but i took this as ultimate tensile strength as i thought it was the same sign for ultimate shear strength