Help engineering hsc MC questions (1 Viewer)

ABBAS38

Member
Joined
Aug 16, 2019
Messages
81
Gender
Male
HSC
2019
Is anyone working it out just let me know pls thx
 

blyatman

Well-Known Member
Joined
Oct 11, 2018
Messages
539
Gender
Undisclosed
HSC
N/A
Max stress is 400 MPa.

Stress = F/A so F = A*stress = (1/4)*pi*(8e-3)^2 * 400e6 = 20,106 N.

So max load it can sustain is 2000 kg.
 

StudyOnly

Active Member
Joined
Oct 16, 2018
Messages
171
Gender
Male
HSC
2019
Let's do this first,
Max stress = 800 MPa = 8 x 10^8 Pa
diameter = 8mm = 0.008m
FoS = 2

We know:
Allowable Stress = Max Stress/FoS
8 X 10^8 / 2 = 4 x 10^8 Pa

Stress = F/A
4 x 10^8 Pa = F/(pi x 0.004^2)
F = 20106.19N

F = ma
20106.19N = m x 10
m = 2000kg (roughly)
 

ABBAS38

Member
Joined
Aug 16, 2019
Messages
81
Gender
Male
HSC
2019
Ok got it but I thought the formula has yield stress in it and max stress not allowable and max allowable. Max allowable stress equals yield/FoS
 

blyatman

Well-Known Member
Joined
Oct 11, 2018
Messages
539
Gender
Undisclosed
HSC
N/A
Max stress is 800 MPa. This is what it was designed to - it will fail if it exceeds this. So a FoS is added in to cap the allowable stress at 400 MPa.

Not sure why you're getting caught up on the yield stress.
 

StudyOnly

Active Member
Joined
Oct 16, 2018
Messages
171
Gender
Male
HSC
2019
Keep this in mind:
You know the formula is stress(allowable) = stress(yield)/FoS for ductile materials

Now stress(yield) is the REAL point of failure, for safety engineers will add a FOS to make sure that the maximum load does not come near to stress(yield) because there is a risk of failure there. As a result they divide by a FOS thus decreasing the maximum (stress(allowable)) so that there is a safety net from failure. In the question it says "A factor of safety of 2 was set by...." which means that that "maximum allowable stress" of 800Mpa is the stress(yield) from the formula. This is why we divide it by 2 to get the stress(allowable).

This is an example of why you should understand what those formulas mean. Use this as a lesson to understand the formulas before the final exams.

I'm sorry if this explanation was crap/confusing but I am in a bit of a rush
 
Last edited:

ABBAS38

Member
Joined
Aug 16, 2019
Messages
81
Gender
Male
HSC
2019
Nah this explanation is really good thanks heaps I get it now!
 

StudyOnly

Active Member
Joined
Oct 16, 2018
Messages
171
Gender
Male
HSC
2019
w = Fs
w = (90cos30 - 30) x 3
= 143.8kJ .......... B

w = Fs
where:
F = force applied in direction (parallel to) of movement
s = distance moved
 

ABBAS38

Member
Joined
Aug 16, 2019
Messages
81
Gender
Male
HSC
2019
w = Fs
w = (90cos30 - 30) x 3
= 143.8kJ .......... B

w = Fs
where:
F = force applied in direction (parallel to) of movement
s = distance moved
I have another question about the pulley I posted it just now
 

ABBAS38

Member
Joined
Aug 16, 2019
Messages
81
Gender
Male
HSC
2019
load = 120 x 10 = 1200N
vr = 4 (cut number of ropes)
ma = load/effort = 1200/400 = 3

Efficiency = (ma/vr) x 100
= 3/4 x 100
75% ....... C
I get it but I don’t really get the velocity ratio part how is there 4 cut ropes and how do I just find them in general for a pulley system(VR)??
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top