ELEMENTARY STRUCTURAL DESIGN
Elementary Structural Design (CIV131) is a module taught to Master of Engineering and Bachelor of Engineering students at Newcastle University. This page is intended primarily for those students. It comprises the module's course notes and other relevant information. Much of the course comprises case studies, which are presented during the 24 lectures. These case studies can be revised by clicking on a topic in the table below. The module is examined by a formal 2 hours examination (80%) and by coursework (20%). To see details of the coursework, click here. When you submit the coursework assignment, you will need to complete a Quality Assurance sheet (one per student). You should use the partcompleted sheet which you will find here. Complete the remainder then attach it to your submission.
The recommended text is:
"Structural Mechanics"
by
Durka, Morgan & Williams
Published by Longman (1996)
ISBN 0582251990
Price £18.99
Available from Blackstones, Haymarket.
Each lecture includes a series of structural case studies which are shown below. Click on a case study to view pictures and explanations:
Click on the cells of the table below to see 10 typical bending/direct force examination questions with their solutions. Click on the left hand column for the question and click on the right hand column for the solution (each solution is on 2 pages, hence two positions to click for each solution).
QUESTION NUMBER 
SOLUTION 
Here are the course notes:
Newton’s Second Law of Motion:
underpins virtually everything a structural engineer does
 although usually our structures don’t accelerate.
Every object has mass. When mass exists in a gravitational field, the body’s mass causes it to have weight, which is a force.
g = 9.81m/sec^{2}
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/mm^{2}, calculate the size of the members in the truss. Select an arrangement of members which can be connected together practically.
KEY POINTS:
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:
R_{A} + R_{B}  20 = 0
Then take moments about A:
R_{B} X 10  20 X 8 = 0
R_{B} = 16kN
R_{A} = 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 (udl’s) are applied e.g. by the selfweight of a beam. Consider the following case:
As always, first calculate reactions at A and B:
R_{A} = R_{B} = 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/2w.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.
It is vital that you become proficient at analysing simple frames and beams since much of the rest of the course assumes that you have this ability.
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 E’F’
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,
where
E = elastic modulus (N/mm^{2})
e = strain (ratio of 2 lengthsdimensionless).
So, stress = 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 beam’s 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 y^{2}
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.y^{2} = E/R.I_{xx}
I_{xx} 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
where
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 = 6830mm^{2} Z = 922.8cm^{3}
2/ 381 x 152 x 52kg/m. Area = 6640mm^{2} Z = 842.3cm^{3}
3/ 305 x 165 x 54kg/m. Area = 6830mm^{2} Z = 751.8cm^{3}
4/ 356 x 171 x 57 kg/m. Area = 7210mm^{2} Z = 894.3cm^{3}
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/mm^{2} in each case.
DEFLEXIONS
It is often the case that we need to know deflexions of structures. A structure can fail because it deflects too much, even if it is strong enough.
We have already shown that M/I = E/R.
R is the radius of curvature of a beam and is given by:
1/R = d^{2}y/dx^{2}
Or,
El.d^{2}y/dx^{2} =M
This means that we can obtain deflexions from a knowledge of the bending moment equation. We need to integrate the bending moment equation twice. Note that this introduces constants of integration which we can obtain by knowing support conditions as follows
Consider a point load at the end of a cantilever
The bending moment at point x is:
BM = W.x  W.L
We can substitute this BM equation into the curvature equation:
Eld^{2}y/dx^{2} = W(Lx)
Eldy/dx = WLx  Wx^{2}/2 + C
When x = 0, dy/dx = 0, so C = 0
Therefore El.dy/dx = WLx  Wx^{2}/2
Integrate a second time:
El.y = Wlx^{2}/2  Wx^{3}/6 + D
When x = 0 then y = 0 so D = 0
Therefore, El.y = Wlx^{2}/2  Wx^{3}/6
The above equation gives the value of deflexion, y at any distance x along the cantilever. In particular, we can find the value of y at the tip of the cantilever by setting x = L so that:
y = WL^{3 }/3EI
Now, consider a simply supported beam
The bending moment is:
M = wLx/2  wx^{2}/2
Therefore, El.d^{2}y/dx^{2} = wLx/2  wx^{2}/2
Integrate once:
El.dy/dx = wLx^{2}/4  wx^{3}/6 + C
When x = L/2, then dy/dx =0, so C = wL^{3}/24
Therefore:
El.dy/dx = wLx^{2}/4  wx^{3}/6  wL^{3}/24
Integrate a second time:
El.y = wLx^{3}/12  wx^{4}/24  wL^{3} x/24 + D
When x = 0, then y = 0 so D = 0
Therefore,
El.y = wLx^{3}/12  wx^{4}/24  wL^{3} x/24
The maximum deflexion at the centre of the beam can be obtained by substituting x = L/2 into the above equation to give:
y = 5wL^{4 }/384EI
Find the equation for the deflected shape of the beam shown.
A simple way of obtaining the deflexion of any point in a frame is now presented. Consider the frame shown above. First, we calculate the lengthening or shortening of each member by first finding the member forces, then applying the relationship E = stress/strain.
Member AC lengthens by v and BC shortens by w. AC rotates round one of the arcs and BC rotates round the other one. Because v and w are small, the movements can be thought of as being perpendicular to each member rather than arcs.
Let E = 200,000N/mm^{2} (steel) and area = 50 mm^{2}.
We can obtain v and w by substituting all the known values as follows:
v = 100000 x 5000/(50 x 200000) = 50mm
w = 160000 x 8000/(50 x 200000) = 128mm
(Note that these are larger than would normally be expected, but they allow us to visualise them better.)
We will now draw the WILLIOT DIAGRAM so as to scale off the true deflexion of point C in the frame.
First, select a joint which does not move as the "pole". If, as was the case in the example, two or more joints do not move, each of them can be considered to be at the pole. Second, draw to scale and in the correct direction, the change in length of each member touching the pole. Third, for each moving joint in a frame, construct a line at right angles to the two members’ vectors which arrive at that joint. These lines intersect at the displaced position of the joint.
SLABS AND PLATES
Slabs spanning in one direction can be designed as if they were wide beams. Slabs spanning in 2 directions are more complex and are dealt with as follows.
A load applied onto the 2 way spanning slab will be transmitted to all 4 sides in such a way that we can visualise the slab as beams spanning in orthogonal directions. In the case of a rectangular 2way slab, most of the load is taken to the longer sides because the shorter beams spanning onto those sides are stiffer. Consider the common case of a slab supporting a uniformly distributed load. We design the slab by calculating bending moments in two orthogonal directions using the following equations:
Mx = a.w.L^{2}_{x} My = b.w.L^{2}_{y}
where Mx is moment acting on beams spanning in the shorter direction.
Lx is the shorter span
The value of a and b depends upon the relative lengths of Lx and Ly and upon the support conditions as in the next table

VALUES OF
1.0 
A FOR LY/
1.25 
LX OF
1.5

VALUES OF B FOR ALL VALUES OF LY/LX 
1 Interior Panels Negative moment Positive moment 
0.031 0.024 
0.044 0.034 
0.053 0.040 
0.032 0.024 
2 One short edge discontinuous Negative moment Positive moment 
0.039 0.029 
0.050 0.038 
0.058 0.043 
0.037 0.028 
3 One long edge discontinuous Negative moment Positive moment 
0.039 0.030 
0.059 0.045 
0.073 0.055 
0.037 0.028 
4 Two adjacent edges discontinuous Negative moment Positive moment 
0.047 0.036 
0.066 0.049 
0.078 0.059 
0.045 0.034 
Note that a negative moment acts at the support and causes hogging. A positive moment acts at midspan and causes sagging.
For Ly/Lx greater than 1.5, use the a values corresponding with 1.5.
EXAMPLE
Design a reinforced concrete floor slab supported on a steel grillage to withstand an imposed load of 5kN/m^{2}. Assume:
concrete density = 24kN/m^{3}
concrete strength = 11N/mm^{2} (or Mpa)
steel strength = 215N/mm^{2}
concrete thickness = 125mm
PanelLx 
Ly 
Ly/Lx 
ap 
an 
bp 
bn 
Mxp 
Mxn 
Myp 
Myn 

A1 
3 
6 

A2 
5 
6 

A3 
4 
6 

B1 
3 
4 

B2 
4 
5 

B3 
4 
6 

C1 
3 
4 

C2 
4 
5 

C3 
4 
6 
Moment of resistance of section = 76 x 170 = 13.9kNm/m
1000
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.
Aquaduct Design
Calculate the bending moment on each beam, assuming each beam carries an equal proportion of the total load.
Design to Resist Impact
Design the posts of a motorway crash barrier such that a post fails in shear when a vehicle of weight 12t impacts at a speed of 3m/sec normal to the barrier. The barrier rail is designed to deform by 1m as the impact occurs. First, determine the force applied to a post by using the equation:
V^{2} = U^{2} + 2fs (don't forget V = U + ft)
Masonry
Design a dry stone wall 1.3m high to resist a horizontal vandal force of 200kg. Consider 1m length of wall.
END