CIVIL 211 STRUCTURAL MECHANICS 2
QUESTIONS AND ANSWERS
SEPTEMBER 2005-09-02
I just have a few questions about some of the assignment questions.
For assignment 1 question 9 do we have to know how to derive the formula for Zpr? and Mpr if it comes up
in another case?
More important to understand what Zpr and Mpr mean and when you might need to use them.
I don't understand what is being asked for in assignment 1 question 10? What do you mean by over what
length of the beam yield stress will have been reached at some point in the cross section?
Yield stress will be reached at every point in the beam where M>=My. My is the 'yield moment', the moment
at which the maximum stress, (M.y/I) just reaches yield. When the mid-span moment reaches MP there will
be a region either side of mid-span where the moment, although less than MP, will be > My.
For assignment 2 question 1 do you have to estimate a self weight and do we need to add up all the live loads
and divide them by the total length of the beam to make it a distributed load for using it in the 1.2G + 1.5Q
equation?
Yes, need to estimate self-weight.
No, the concentrated live loads should remain as concentrated loads (and be multiplied by 1.5). The
distributed dead load (12kN/m + self-weight) should remain as a distributed load (and be multiplied by 1.2).
Loading then consists of a uniformly distributed load together with the three point loads.
Thanks, that was really helpful, although I'm not completely certain about the sway mechanism in the
"angled" portal frame. I'll have to come see you sometime.
I've got a question regarding some stuff in the Assignment 2 solution sheet.
for question 1, you've got
1.2G + 1.5Q=15.25kN/m......how is that possible? then you've got +300,225,150KN........how did you get
these values? the point loads are 200,150,100KN which are the live loads.
The point loads of 200,150,100 are live loads and therefore need to be multiplied by 1.5.
The spread load of 12.7kN/m is the dead load, therefore mult by 1.2.
Also for you're mechanism calculations you've used the "new" points loads, 300,225 and 150.......why?
Objective is for the structure to just collapse under the factored loads (1.2G+1.5Q). Hence the factored loads
are used in the collapse load calculation.
And for mechanism A and B calculation, you've written 15.25x 6x2 x theta......how did you get this?
why 2 theta? and where is the "lever" arm for the distributed load?
15.25kN/m x 6m is the total spread load on the beam. 2m x theta is the mean vertical displacement of this
load. Product of total load x mean displacement gives (virtual) work done.
The next bit I'm a little unsure about is you're elastic analysis of question 1.
For FEMs: Mab = -15.25 x 6^2/ 12 - (300x2x4^2/6^2) ( what formula did you use for this?)
WL^2/12 is for the spread load.
Point loads are considered one at a time and superimposed using the formula:
Beam with point load of P at distance ‘a’ from left hand end, A, and ‘b’ from right hand end, B.
MAB = Pab^2/L^2, and MBA = Pa^2b/L^2.
I've got some questions regarding page 20 of the plastic analysis hand out.
For the Beam mechanism, you've written the eqn.
Mp(column)(2 theta)+ Mp(beam)(2 theta)= 8W.6.theta
now, wouldn't the column be = 0? as there is no rotation in horizontal direction.
Rotations are about an axis perpendicular to the plane of the structure (not in "x direction").
The angle between the column and the beam (90 degrees initially) changes by theta - this is the hinge
rotation.
And for combined mechanism. You've got:
Mp(column) (4 theta) + Mp(beam) (2theta)= .........
now, wouldn't the column angle be 3 theta instead for 4 theta as there is no plastic hinge at hinge location 2?
I agree that there is no plastic hinge at 2. However, there are plastic hinge rotations at 1 (1 x theta), 4 (2 x
theta) and 5 (1 x theta): TOTAL = 4 x theta!
And, why are you using the beam mechanism to find what M2 is? its the combined mechanism that gave the
lowest load. so what not use combine then?
The beam mechanism is used merely to express the equilibrium relationship between the 8W vertical load
and the moments in the beam at points 2, 3 and 4. Since the moments at 3 and 4 are known (Mp of beam and
column respectively) the only unknown in the equilibrium equation is M2.
If the combined mechanism was used it would give us an equilibrium relationship between the Mp values at
1, 3, 4 and 5 and the 6W and 8W loads, leading to W = 3.889. But we've already done that to get what we
hope is the right answer. Also it doesn't tell us what the moment at 2 is anyway because there is no hinge at
2.
thanks. I'm looking forward to your reply. For the test on the week coming back, what do need to know?
plastic analysis and buckling? or just plastic analysis.
Just plastic analysis.