Explain with reference to the 'standard equation of bending' why it is possible to remove material from the vertical element of an 'I' section beam without significantly reducing its load bearing capabilities.

Question 2

                A beam ABCD is simply supported at A and D and is 6 metres long. It carries concentrated loads of 48kN and 40kN at B and C respectively. If AB=1m and BC=2m, using Macaulay’s method find the deflection under each load and the position and magnitude of the maximum deflection. E and I are 200GN/m2 and 85 x 10-6m4 respectively.

Answer 2

Fig. 1 – Drawing of problem

First, take moments about Ra,

CW=CCW

So,

Therefore,

Rb=28kN

& Ra=60kN.

Taking bending moments about XX,

Therefore,

By integration,

And,

Considering what is happening on the beam,

• When x=0 then y=0 then B=0
• When x=6 then y=0 and  (found by substituting 6 for x in EIy equation)

So,

……(i)

and,

……(ii)

To find the deflection under each of the point loads the second equation is used and the distances from Ra are placed into it in place of x.

So,

when x=1

It is raised by 103 because the rest of the calculation is worked in kilonewtons.

Therefore,

When x=3

So,

Now the deflections under each load have been determined the maximum deflection and its location along the beam can now be calculated. So,

To find point of maximum deflection use equation (i) and let =0.

 

since, by inspection x must be less than 3

Expand the brackets,

The quadratic formula can now be applied to obtain the distance from Ra to the point of maximum deflection. So when,

therefore,

Since the point of maximum deflection cannot be off the bar it can only be located at 2.8718metres from Ra. All that is left is to find the maximum deflection. So,

So, to conclude question 2, the deflection under the 48kN and 40kN loads are –9.019mm and –16.7mm respectively. The maximum deflection is –16.7459mm at 2.8718m from Ra.

Question 3

                A beam ABCD is simply supported at A and D and is 8 metres long. It carries a UDL of 8kN/m over its entire length. Using Macaulay’s method determine the magnitude and position of the maximum slope and the maximum deflection that is present on the beam. EI=700kN/m2.

Answer 3

Fig. 2 – Drawing of problem

First convert the UDL to a point load.

This will act in the middle of the beam as it is the only load to be considered. This also gives 32kN of reaction at Ra and Rb.

Therefore,

So, by integration,

And,

Let us now consider what is happening on the beam,

• When x=0 then y=0 then B=0
• When x=8 then y=0 and  (found by substituting 8 for x in EIy equation)

Therefore, maximum deflection is,

The slopes of the beam may now be considered. The maximum slope occurs when x=0 and x=8. So,

When x=0

When x=8

So, to conclude question 3, the maximum slope is at Ra and Rb respectively and the maximum deflection is –609.523mm at the middle of the beam.

Question 4

                A cantilever beam is 2m long and carries a UDL of 15kN/m over its entire length together with a concentrated load of 8.46kN 1.5metres from the fixed end. If the beam is 75mm wide, 150mm deep and made from material with an E value of 210GN/m2, determine the deflection of the free end.

Answer 4

Fig. 3 – Drawing of problem

First to consider is the bending moments about XX.

So,

Therefore, by integration,

Let us now consider what is happening on the beam,

• When x=2 then y=0
• When x=2 then =0

So, when x=2, y=0 and

and when x=2, =0 and

Since,

So,

Now to calculate I.

Now EI may be calculated, hence

Finally the deflection may now be calculated.

So, the deflection of the free end when under the prescribed load is –9.995mm.

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