A beam 2.4 m long is simply supported at two points A and B which are 1.44 m apart; point A is 0.36 m from the left-hand end of the beam and point B is 0.6 m from the right-hand end; the value of EI for the beam is 240×10^8 Nmm^2. Find the slope at the supports due to a load of 2 kN applied at the

Question

A beam 2.4 m long is simply supported at two points A and B which are 1.44 m apart; point A is 0.36 m from the left-hand end of the beam and point B is 0.6 m from the right-hand end; the value of EI for the beam is [latex]240 imes 10^{8} Nmm ^{2}[/latex]. Find the slope at the supports due to a load of 2 kN applied at the mid-point of AB.
Use the reciprocal theorem in conjunction with the above result to find the deflection at the mid-point of AB due to loads of 3 kN applied at each end of the beam.
Ans. 0.011, 15.8 mm.

Accepted Answer

Referring to Fig. S.15.25

the support reactions are each equal to 1 kN from symmetry. The slope of the beam at the supports is given by Eq. (iii) of Ex. 13.5

[latex]E I \frac{ d v}{ d x}=\frac{W}{16}\left(4 x^{2}-L^{2} ight)[/latex] Eq. (iii) of Ex. 13.5

and is

[latex]\frac{ d v}{ d x}=\pm \frac{2 imes 10^{3} imes\left(1.44 imes 10^{3} ight)^{2}}{16 imes 240 imes 10^{8}}=\pm 0.011[/latex]

The deflection at C is then [latex]0.011 imes 0.36 imes 10^{3}=3.96 mm[/latex] and the deflection at D is [latex]0.011 imes 0.6 imes 10^{3}=6.6 mm[/latex] .
From the reciprocal theorem the deflection at F due to a load of 3 kN at C = 3.96 × 3.0/ 2.0 = 5.94 mm and the deflection at F due to a load of 3 kN at D = 6.6 × 3.0/2.0 = 9.9 mm.
Therefore the total deflection at F due to loads of 3 kN acting simultaneously at C and D is 5.94 + 9.9 = 15.84 mm.

Question 15.P.25: A beam 2.4 m long is simply supported at two points A and B ...

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