Recent content by ridiculoid

Homework Statement A sample is put in tension and a strain rosette gives the following results: i) Calculate principal strains and poissons ratio using Mohrs strain circle. ii) Calculate principal stresses from principal strains and poissons ratio. Homework Equations Mohr's strain circle γ =...

Homework Statement Homework Equations Trigonometry equations, p=m/v, A = pi * d^2 / 4 The Attempt at a Solution i) mass calculated is 9.318kg at 150° from A ii) diameter = 0.235m I apologize for the messy working, I am just wondering if I have followed the correct procedure or not...

Ahh that makes sense! Thank you for the help.

Homework Statement [/B] problem - https://imgur.com/34yTyCAHomework Equations I = d . b^3 / 12 tmax = (F / I . b) . (A1 . y1) + (A2 . y2)The Attempt at a Solution It is asking for the maximum shear FORCE in the beam. Do I take the maximum value from the shear force diagram as shown in my...

Homework Statement [/B] problem - https://imgur.com/SKeTUXNHomework Equations Mohrs circle The Attempt at a Solution [/B] I have drawn 2 induced loading elements (shown at the bottom of of my working) I just seem to have conflicting information from some class examples and am wondering...

Homework Statement Mohrs circle question - https://imgur.com/SKeTUXN Shear stress in I beam question - https://imgur.com/34yTyCAHomework Equations Mohrs Circle - None Shear stress in I beam - I = d . b^3 / 12 tmax = (F / I . b) . (A1 . y1) + (A2 . y2)The Attempt at a Solution For the...

Great, thanks so much for the help. Got there in the end hahah. The doubled value probably comes from the fact the pins are under double shear, 2 pins with 2 shear points each puts the shear value for the pins at force = stress * 4 * Area. That's how I interpreted it anyway, It still comes in at...

Right, I should say force = stress * area. When I calculated P around the maximum force that can be applied to the brass (4.8MN) and steel pins (814.3kN) I got P as 13.8MN and 1.25MN respectively, which by this point the steel rod will have failed. So to clarify, am I going down the right...

Seeing as the metal rod has a yield stress of 640MPa, I used stress = force / area to find the the amount of force before yield which is 201kN? I used the formula on the other components as well as since the steel rod yeilded at the lowest force I thought I should start there.. I am not sure...

Right, so I take moments about the 201.06kN force and find the reactive force at the brass block is 107kN which sums to P being 308kN. Does this seem correct? Am I right in taking the maximum force of the steel rod (201.06kN) as my left reactive force as this will fail before the pins? (814.3kN)...

Thanks for the response, I understand what you're saying, will the force P be distributed between the pins / rods and block up to 201kN? I'm just trying to figure out a solid value to work from. Am I to multiply the force at the pin by 0.8 for the moment to the left of P? I'm still confused as...

Homework Statement See attached photo Homework Equations Stress = F/A Area = pi d^3 / 4 Sum F h = 0, Sum F v = 0, Sum M = 0 The Attempt at a Solution Hey everyone, after reading the question a number of times and not knowing for certain where to start, I figured because the stress and area...

Good point, they are 2 identical pieces overlapping and pinned together at the shoulder

Ah yes that does make sense, and I see where I've calculated that wrong. Thanks very much for your help.

By arbitrary I mean just counting the holes in the bracket as units, not using the measurements which I completely forgot I had taken, feel pretty stupid.