Applied Engineering Technology – Level 4
A look at simple beams
A possible question on beams can be as follows:
Use the following information and draw the beam to the scale 1 meter = 1 cm.
A beam of 12 meter is supported at L and R. Point L is 3 meter from the left end of the beam and point R is 2 meter from the right end of the beam. There is a point load of 40N on the left end of the beam and a point load of 20N on the right end of the beam. There is another point load of 50N, 3 meter from the right end of the beam. A uniformly distributed load of 10N/meter starts at L and end at the point load of 5N. Ignore the mass of the beam.
Draw a simplified diagram to scale to solve the reactions at L and R
At first glance this problem seem to be massive because of all the confusing information, but remember that you can eat a big elephant one bite at a time. So let us attack this problem one step at a time.
Step 1: A beam of 12 meter – We stop here and draw a line 12 cm long (scale 1meter = 1 cm.)
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Step 2: Point L is 3 meter from the left end of the beam. We measure 3 cm from the left end to find L
Step 3: Point R is 2 meter from the right end of the beam. We measure 2 cm from the right end to find R
Step 4: There is a point load of 40N on the left end of the beam.
Step 5: There is a point load of 20N on the right end of the beam.
Step 6: There is another point load of 50N, 3 meter from the right end of the beam.
Step 7: A uniformly distributed load of 10N/meter starts at L and end at the point load of 5N.
And that is your final drawing, your first elephant has been eaten. Now we must start to work out the forces at L and R.
First we need to transform the UDL (uniformly distributed load) of 10N/m to a point load. It is important to work out the length over which the load is distributed. The load starts 3 meter from the one end of the 12 meter beam and ends 3 meter from the other end. Therefore we are going to take the length of the beam and subtrack the the two 3 meter ends. The calculation will look like this. 12-(3+3) = 12-6 = 6 meters.
Now we know that the UDL is distributed over 6 meters at a load of 10N/m (given) The pointload will thus be 6 X 10 = 60N and it will be in the centre of the UDL. Let’s make it part of our drawing:
Let us start with the value of L. – To work out the value of L we must determine the movements around the centre point R. remember that the whole beam can now rotate around the point of R. If we count all the forces we find the following from left to right: 40N , L , 60N , 50N and 20N. Note that we do not count R. So there are 5 forces and we must make sure that we take all of them into account.
Now we must decide which of these five forces will work clockwise around R and which of them will work counter-clockwise. This is easy if we follow the following rule of thumb:
All downward forces to the left of the point R will move counter-clockwise.
All downward forces to the right of the point R will move clockwise.
The opposite is true for the upward forces.
All upward forces to the left of the point R will move clockwise.
All upward forces to the right of the point R will move counter clockwise.
Let us start with the force at L because we want L to be on the left hand side of our equation.
You will find that L is an upward force on the lefthand side of the centre at R and moves the beam in a clockwise direction around R. To find the downward forces that will move the beam clockwise around R (the same direction as L) we must look for downward forces to the right of R. There is only one downward force to the right of R and that is the 20N force ar the end of the beam. Logic now dictate that the remaining three forces wil be counter-clocwice around R because they are downward and to the left of R.
Each force stand in relation to the distance from R and the clokcwise forces must be equel to the counter-clockwise forces. We have two clockwise forces = three counter-clockwise forces. The Equation wil be:
(force X distance)+ (force X distance) = (force X distance)+ (force X distance)+ (force X distance)
(L X 7m)+(20N X 2m) = (40N X 10m)+(60N X 4m)+(50N X 1m) Make sure your distances from R is correct.
(L X 7m)+ 40 = 400 + 240 + 50
(L X 7) = 690 – 40
L = 650 ÷ 7
L = 92.86N
We worked out L. Now we must do the same for R. To work out the value of R we must determine the movements around the centre point L. Remember that the whole beam can now rotate around the point of L. If we count all the forces we find the following from left to right: 40N , 60N , 50N , R and 20N. Note that we do not count L. So there are 5 forces and we must make sure that we take all of them into account.
Now we must decide which of these five forces will work clockwise around L and which of them will work counter-clockwise. This is easy if we follow the following rule of thumb:
All downward forces to the left of the point L will move counter-clockwise.
All downward forces to the right of the point L will move clockwise.
The opposite is true for the upward forces.
All upward forces to the left of the point L will move clockwise.
All upward forces to the right of the point L will move counter clockwise.
Let us start with the force at R because we now want R to be on the left hand side of our equation.
You will find that R is an upward force on the righthand side of the centre at L and moves the beam in a counter-clockwise direction around L. To find the downward forces that will move the beam Counter-clockwise around L (the same direction as R) we must look for downward forces to the left of L. There is only one downward force to the left of L and that is the 40N force ar the end of the beam. Logic now dictate that the remaining three forces wil be clocwice around L because they are downward and to the right of L.
Each force stand in relation to the distance from L and the clokcwise forces must be equel to the counter-clockwise forces. We have two counter-clockwise forces = three clockwise forces. The Equation wil be:
(force X distance)+ (force X distance) = (force X distance)+ (force X distance)+ (force X distance)
(R X 7m)+(40N X 3m) = (60N X 3m)+(50N X 6m)+(20N X 9m) Make sure your distances from L is correct.
(R X 7m)+ 120 = 180 + 300 + 180
(R X 7) = 660 – 120
R = 540 ÷ 7
R = 77.14N
Well done! We have calculated the forces at L and R. Now to make sure thet we have done so correctly. If the forces at L and R is correct the sum of the two upward forces should be equal to the sum of all the downward forces.
L + R = F1 + F2 + F3 + F4
92.86 + 77.14 = 40 + 60 + 50 + 20
170 = 170 And they balance so our calculations were correct.
Now, let us work out the Shear force diagram: We do this directly underneathour drawing that we started with. Draw very light construction lines down from every force (up or down force) and then draw a zero line.
Our Shear force diagram must now be build around the zero line. We will work from left to right and start on the left side of the zero line. Our first force is 40N downward. Strarting from zero we go four centimeter down to A.
For the next three meters there are no force on the beam. We can therefore draw a line straight across to B.
At B we find an upward force of 92.86N (Force at L) Starting at B we draw a line of 9.3cm straight up to C.
From here we find a downward UDL (Unifomed Distributed Load) of 60N over the next 6 meters. We measure 6cm across to D (6 meters) and 6cm down to E (60N). Now we draw a diagonal line from C to E. A UDL will always form a diagonal line.
At point E there is another downward force of 5N so from E we draw a 5cm line down to F.
From F to G there is no force. We draw a 1cm line from F to G.
At G we find that the upward force at R apply. That tells to draw a line of 7.7cm straight up from G to H.
From H to I there is no force. We draw a 2cm line from H to I.
The downward force of 20N at I turn into a 2 cm line that will end on the zero l
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