Beams are the backbones or the ribs of every construction or structure. Therefore calculating beams to their optimal conditions is absolutely critical and becomes the prime necessity for all constructions. However along with the beams, the structires supporting them also become crucial as these are responsible for keeping the structure erect and standing. The following two examplea clearly illustrate the practical situations and the remedies that needs to be heeded upon while installing such constructions.
Calculating Beam Support: Consider a beam that needs to carry loads up to 500 kg including its own weight and suppose it is to be built over two supports through its upper end and the lower ground ends.
Also consider the given inclination of the construction is at an angle of 30 degrees, the length and the height of the being are 400 and 50 centimeters respectively.
For such a construction it would become a prior necessity to calculate the amount of average weights the two ends would be subjected to.
Let’s calculate the average magnitudes of weight that the upper and lower ends would be required to carry.
Here we have,
Length = 400 cm,
Height = 50 cm,
Angle of Inclination, ? = 30 degree
Weight = 100 Kg.
Let’s denote the above given magnitudes with reference to the figure as:
P = Weight subjected at point A
Q = Weight subjected at point B
C = the point over the lower surface from where the center of gravity cuts through the beam.
Also let’s join G with M, where G is the center of the force of gravity and M is the center of the lower plane of the beam.
The shape or the geometry of the design suggests the following expressions:
GM = 50/2 = 25 cm,
AM = 400/2 = 200 cm,
AC = AM – CM = 200 – tan 30 degree, because CM/GM = tan CGM = tan 30 degree
= 200 – 50/√3 = 29 cm.
And therefore CB = AB – AC = 400 – 29 = 371 cm
Using the standard expression, P × CA = Q × CB, we get
P × 29 = (500 – P) × 371, because P + Q = 500 kg
P × 29 = 185500 – 371P
29P + 371P = 18550,
400P = 18550,
P = 18550/400 = 46.375 Kg
And Q = 500 – 46.37 = 453.63
Calculating Beam Support with Cords: Consider a beam with a uniform 3 meter length, weighs 40 kg, is suspended horizontally by a couple of vertical cords. The cords have the strength of sustaining not more than 35 kg of load.
With the above conditions let’s try to assume a situation where one of the cords just snaps when a 20 kg weight is placed over the beam. Here the position of the 20 kg weight is critical and we’ll to calculate how this weight may be position so that one of the cords just gives away.
The given data are:
Length of the beam = 3 m,
Weight of the beam = 40 kg,
Load or the tension handling capacity of the individual cords is = 35 kg
Weight of the external body to be placed over the beam = 20 kg.
Suppose the weight of 20 kg is placed somewhere at point D at a distance x from A, over the beam.
A little thought and analysis shows that one of the cords may just break if the load over it exceeds 35 kg. In other words, the cord breaks when the moment of force over it crosses 35 kg, which may be represented as R(A) = 35 kg.
Taking moments about the point B and equating gives:
R(A) × 3 = 20 (3 – x) + 40 × 1.5
35 × 3 = 60 – 20x + 60
20z = 120 – 105 = 15
x = 15/20 = 0.75m
Therefore, the above result informs us that the load of 20 kg will break the cord at A if it’s placed at a distance of 0.75 meters from the point A and will brak the cord at B, if the load is placed at a point 0.75 meters from the side B.
Why don’t you try solving following problem, with reference to the above solved condition?
Consider a rod with a length of 80 cm; virtually having no weight by itself is supported in the horizontal position across the points A and B. Suppose a load of 30 kg is held somewhere at point C over the rod, then find the position of the point C from the side A required to break the support at A, assuming the tolerance of the support at A is not more than 12 kg.