STRUCTURES III FOR ARCHITECTS 2002-2003
A local newspaper "The Sunday Sun" (not to be confused with "The Sun") published an article following the installation of the Newcastle Millennium Arch Bridge in which they compared several historic bridges in the NE. I would like you to read the article as part of this course. Click here to read it.
Click here to go to Gateshead's Millennium Bridge site.
The recommended text for this module is:
Durka, Morgan & Williams
Published by Longman (1996)
Available now from Blackstones, Haymarket.
Solutions to the questions in the 2000 Structures III exam can be seen by clicking on pages in the table below. You should all have a copy of the exam - it was a handout at the start of the module.
Page 1 (Q1)
Page 2 (Q2)
Page 3 (Q2 contd.)
Page 4 (Q3)
Page 5 (Q3 contd. and Q4)
Page 6 (Q4 contd. and Q5)
Page 7 (Q5 contd.)
Page 8 (Q6)
Page 9 (Q7)
Solutions to the 2001 exam can be seen by clicking on the pages in the table below.
INTRODUCTION TO MODULE
Structures III is a 5 credit module taken by Stage 3 Architects at Newcastle University. The theoretical material is set out in these notes. In addition, case studies are used to apply the theoretical material. Click on case study subjects in the table below.
All lectures take place at 11am on Friday mornings in Cassie Building Room 315 on the third floor near the front of the building - the front is near the Claremont Road pedestrian crossing. The timetable is shown below. Case studies will be presented in the lectures as follows.
Lecture 1 - Trusses. 27th September
Lecture 2 -Trusses. 4th October
Lecture 3 - Trusses. 11th October
Lecture 4 - Steel. 18th October
Lecture 5 - Suspension Bridges. 25th October
Lecture 6 - Arches. 1st November
Lecture 7 - Arches. 9th November
Lecture 8 - Refurbishment. 15th November
Lecture 9 - Industrial Building. 22nd November
Lecture 10 - Bending. 29th November
Lecture 11 - Bending. 6th December
Lecture 12 - Special Bridges. 13th December
Click here for details of the assignment which carries 50% of the assessment of this 5 credit module
Before looking at the technical notes, here are some supplementary diagrams and pictures which we will refer to in lectures. First, here is the comparison between a truss, a cantilever and an arch for the relatively large 100m span, carrying typical highway loading.
Below is the way in which the Swing Bridge works when in the open and closed positions. Note that when the bridge is open to road traffic, the self weight of the bridge is taken as shown, with the bridge acting as a single span beam, but the imposed load (traffic) is supported through the pivot support at the centre of the bridge.
The picture below, taken in late 1928 shows how the partly completed arch of the Tyne Bridge was supported during its construction.
Click here for details of structural steel sections
Now, here are the notes:
Newtons Second Law of Motion:
underpins virtually everything a structural engineer does
- although usually our structures dont accelerate.
Every object has mass. When mass exists in a gravitational field, the bodys mass causes it to have weight, which is a force.
g = 9.81m/sec2
If the surface disappeared, the block would accelerate at that value. Usually, the first thing we need to do when analysing a structure is calculate reactions - without reactions, there would be no forces in structures.
Because the truss is symmetrical with the load at the centre, each end reaction must be 5kN. Therefore, a structural engineer sees this structure as in the next diagram as far as the structure is concerned, there is no difference between loads and reactions.
Suppose the truss were to be fabricated from timber which can be allowed to attain a stress of 5N/mm2, calculate the size of the members in the truss. Select an arrangement of members which can be connected together practically.
1 Calculate reactions first
2. Use consistent sign convention
3. Compression members - arrows point out
Let us now calculate reactions when the structure or the load is not symmetrically disposed.
First, resolve vertically:
RA + RB - 20 = 0
Then take moments about A:
RB X 10 - 20 X 8 = 0
RB = 16kN
RA = 4kN
Now try this:
Because the loads were applied at the nodes, none of the members were subjected to bending only tension or compression occurred in the members. Let's see what happens when something bends:
Even though the beam changes shape, because the change is usually small, we can assume that it is a straight line. Instead of direct tension or compression, The beam is now stressed by a Bending Moment.
To calculate the bending moment at any point x along the beam, cover everything to one side of x and multiply all the forces which you can still see by their perpendicular distance to x:
When x<L/2, BM = Wx/2
When x>L/2, BM = Wx/2 - (x - L/2)W
Find the BM when x = 0,L/8, L/4, 3L/8, L/2, 5L/8, 3L/4, 7L/8 and L. Draw the bending moment diagram.
So far, all of the loads have been point loads. Often, uniformly distributed loads (udls) are applied e.g. by the self-weight of a beam.
Consider the following case:
As always, first calculate reactions at A and B:
RA = RB = wL/2
Cover part of the structure as before and find the moment of all the forces to one side of the point being considered (x).
BM =wLx/2-w.x.x/2 at any point x
Calculate the BM when x =0, L/2 and L. Draw the bending moment diagram. What are the maximum bending moments for a point load at the centre of the beam and for a udl?
Now try this:
Now try these, remembering always to calculate reactions first, then cover part of the structure. In each case, draw the shear force diagram and the bending moment diagram.
In each case, draw to an exaggerated scale the shape of the deflected structure. Bear in mind that the curvature is proportional to the value of the bending moment and when the bending moment is zero, the member is straight - we call this a point of contraflexure.
In the case of direct forces, once we knew the force in a member, it was an easy matter to calculate the stress. In the case of bending, we need to examine the stress distribution inside the beam. Consider a beam initially straight subjected to bending moment.
The moment M causes this short section of beam to bend through a small angle da. Fibre EF is in tension and grows to EF
Strain is defined as extension divided by original length. Therefore, the strain of EF is:
(E'F' - EF)/EF = ((R + y)da - R.da)/(R.da)
So: strain = y/R
Which means that strain is proportional to distance from neutral axis.
Stress is defined as E.e,
E = elastic modulus (N/mm2)
e = strain (ratio of 2 lengths-dimensionless).
So, stress, s = E.e =E.y/R from previous sheet
This gives us the important relationship:
s/y = E/R
i.e. stress is linearly proportional to distance from neutral axis.
The beam resists externally applied M by an internal moment which can be thought of as the force in each fibre multiplied by the distance of that fibre from the beams neutral axis.
The contribution of EF to the internal moment of resistance is:
Stress x Area of EF x y
We can use the equation above to eliminate stress from the above:
E/R x Area of EF x y2
Summing the moment of resistance contributed by each fibre of the beam gives the total moment of resistance which has to be equal to the externally applied moment M.
Therefore, M = (E/R) S area.y2 = E/R.Ixx
Ixx is the second moment of area of the beam
i.e. s/y = M/I = E/R
We sometimes introduce the term Section Modulus which is I/y so:
s = M/Z
Z is the section modulus
Now, select steel members for the following frame from the Universal Beams listed below:
1/ 406 x 178 x 54kg/m. Area = 6830mm2 Z = 922.8cm3
2/ 381 x 152 x 52kg/m. Area = 6640mm2 Z = 842.3cm3
3/ 305 x 165 x 54kg/m. Area = 6830mm2 Z = 751.8cm3
4/ 356 x 171 x 57 kg/m. Area = 7210mm2 Z = 894.3cm3
First, find the reactions, then take moments at C of everything to one side of C to obtain the value of the horizontal force H at A and E. Then draw the BMD, obtain the maximum moment which each member needs to resist and take account of the compressive force in each column. The allowable bending and compressive stresses in the steel are 155N/mm2 in each case.
Moment of resistance of section = 76 x 170 = 13.9kN-m/m
This section is satisfactory. Although we could try to reduce slab thickness in other places, we would normally retain a constant thickness and change steel sizes - but only in a manner which did not make the floor unbuildable. Minimum cost is far more important than minimum weight.
Calculate the bending moment on each beam, assuming each beam carries an equal proportion of the total load.