How do I calculate the factor of safety for a prismatic bar under tension?

AI Thread Summary
To calculate the factor of safety for a prismatic bar under tension, the correct formula is the allowable stress divided by the actual stress. Given a load of 37 N and a cross-sectional area of 5.7 cm² (0.00057 m²), the actual stress is calculated as 37 N / 0.00057 m², resulting in approximately 64,912.28 Pa or 64.91 kPa. The factor of safety is then found by dividing the allowable stress of 100 kPa by the actual stress of 64.91 kPa, yielding a factor of safety of about 1.54. The confusion arises from miscalculating the actual stress and misunderstanding the formula for the factor of safety. Understanding these calculations is essential for accurate engineering assessments.
Kazaam
Messages
1
Reaction score
0
There is a question on my home work that I just don't understand how they got the answer they did.

The first question is:
In a prismatic bar, the load applied on both ends is 37 N causing the member to be in tension. If the cross section area of the member is 5.7 cm2 and the allowable stress is 100 kPa. What is the factor of safety for the system?

I got .1.298 kpa, the answer is supposed to be 1.541 kpa, how does (37N/.057)/100 kpa= 1.541 Kpa, what am I missing here?
 
Last edited:
Physics news on Phys.org
Kazaam said:
There is a question on my home work that I just don't understand how they got the answer they did.

The first question is:
In a prismatic bar, the load applied on both ends is 37 N causing the member to be in tension. If the cross section area of the member is 5.7 cm2 and the allowable stress is 100 kPa. What is the factor of safety for the system?

I got .1.298 kpa, the answer is supposed to be 1.541 kpa, how does (37N/.057)/100 kpa= 1.541 Kpa, what am I missing here?
5.7 cm^2 = 0.00057 m^2. And then you incorrectly have written that F.S. = actual stress/allowable stress...what should that equation be??. And what are the units of the safety factor?
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top