The beam below carries a load of triangular distribution with a maximum value of 3kN/m at midspan. The beam is made of Glulam with E=16GPa and has a second moment of area Imajor =1×10^9 mm4
Determine the deflection at mid-span due to the applied loads.
HINT (1) Determine the reactions due to applied loads (2) Determine the equation of the moment diagram. ( 3) use a unit load
(4) 1GPa is 1000N/mm2
Solution coming soon
Here is question for all of you retaking Structures 2 in September 2013
Imagine you are assessing the interesting cantilever beam arrangement shown below, you have conclude that the moment capacity of the supports is adequate but you are worried about the overall deflection at the very end of the structure. Both cantilever beams are of the same length and have the same EI value of 20 MN.m2. - ( sorry i cant get wordpress to make squared)
By formulating a suitable compatibility equation and through using Flexibility analysis determine the deflection at C.
This is a bit easier than a standard exam question but would carry about 50-60 % of the available marks.
HINTS coming soon!
- Firstly read through the solution for Q4! It uses the same analysis technique.
- Draw yourself a primary structure and a Unit load structure
Judging from last weeks performance quite a few of you have forgotten how to draw BMD. So the next few questions will focus on this issue.
The question focusses on a 3 span bridge with a drop-in span in the middle similar to the Ness Bridge in Inverness Scotland, check out the Happy pontist blog: a great resource from civil engineering students.
Please identify the INCORRECT BMDs
HINT. The trick with such 'drop-in' structures is that the double pin in the middle isolates the effects of load ( moment) from either side. That is to say that a load on the first span cannot produce any moment on the middle or third span.UPDATE 15/7/13. Judging from the number of incorrect answers this question has received I will be leaving it up for a couple of days more.
If you draw the deflected shape for the case where there load is at any point between the first support and the adjacent pin, you should see that the middle span(between pins ) is always straight just rotated slightly)
Try the question again and attempt to draw the deflected shapes.
SOLUTION: 17/6/13. A,C,D are the INCORRECT BMDs (this is what you were asked to submit) the rest of them (B,E and F) are the correct BMD for the structure.
Here is a question from structures 2 but i'm sure first years will get it right.
Question: Which BMD is the correct for the cantilever beam shown?
Solution: Lets start from the point where the load is applied. It might help if you sketch the problem and mark a line across the structure showing the direction of the applied load
- The bending moment will increases linearly when we move away from the load ( we are moving left) - all diagrams show this!
- As we move down the vertical member on the left hand-side the moment is constant as we are always the same perpendicular distance from the direction of the load.
- Note that at the corner the BMD are both on the outside as there is tension on this face.
- As we move along the bottom beam ( left to midpoint) we get closer and closer to the direction of the applied load so the moment decreases linearly
- As we go past the midpoint on the bottom beam the bottom beam we get further away form the direction of the applied load and hence the moment increases linearly-only diagrams C & D show this.
- If we move up the vertical member on the right hand side we stay a constant perpendicular distance form the direction of load and so the moment is constant - the moment is on the inside of the member as this is the face where we have tension. – Only diagram D shows this.
Lastly as we move along the top beam (right to left) the moment must become less negative and reach zero at the point where we are inline with the direction of load, when we have travelled past this towards the support the moment the moment increases and becomes positive indicating a region of hogging bending ( tension on top face) near the support
Okay, this is definitely a simple question to start with.
Question: If x= L/3 what is the vertical reaction at support A?
Solution: To solve this we need to take moment about a point that we know the moment for. Point B is a pin support for must have zero moment.
- So taking moments about B we have acting clockwise the moment generated by the vertical reaction A (Va * L).
- This must be equal to the anticlockwise moments generated by the applied loads, which are equal to
- P*(2L/3) for the moment from the point load
- (wL/2)*(L/4) for the uniform loading - the first term in brackets is the total load the second term is the lever arm to the centre of the load.
3. equating clockwise and anticlockwise and dividing by L we get Va=(2P/3) +(wL/8)
The multispan bridge shown carries three different load types, Dead load (Red), Super-imposed dead load (Blue) and Live load (Yellow). The individual spans are identical in length and stiffness.
Which plausible load case will produce the maximum hogging moment in the structure?
Which plausible load case will produce the maximum sagging moment in the structure?
If you are struggling with this then have a look at the Solution, it involves influence lines
Something for my Bridge Engineering Students
Well done to the 62 correct submissions, below are the solutions for those of you ( not very many) who got it wrong.
Moment at B is Diagram 3, when the load is between B and D the moment at B is always zero
Moment at C is Diagram 4, when the load is at A, the moment C is hogging
Reaction at B is Diagram 2, when the load is at A the reaction at B is greater than when the load is placed directly on B
Reaction at D is diagram 1, when the load is at A the reaction at D is downwards, as the load move between B and D the reaction at D acts upwards with increasing magnitude.
Something for all my Bridge Engineering students.
Pre-stressing a simply supported beam
For the simply supported pre-stressed beam shown above, calculate the maximum allowable length (in meters) if the maximum tensile stresses due to the unfactored dead weight are limited to 1N/mm2.
Take the density of concrete to be 24kN/m3 and the centroidal axis is at mid-height of the section.Please ignore any reliving effects or the action of live loads
Hopefully you got an answer of 36.7m, which is less than the transportable limit of 40m. If you got anything other than 36.7m please read the solution. week13 solution
For the structure shown below calculate the position of the roller support from the free end of the beam (C) such that the moment at the fixed end (A) is zero kNm.
If you didnt get it right or were unsure of the BMD or deflected shape check out this SOLUTION