The displacements at three points, 1, 2, and 3 on a beam were measured when loads were applied separately at each point; the flexural rigidity of the beam is EI. Initially, when a load W1 = 400 N was applied at the point 1 the displacements were Δ1 = 0.75 mm, Δ2 = -0.25 mm, and Δ3 = -1.25 mm. Find

Question

The displacements at three points, 1, 2, and 3 on a beam were measured when loads were applied separately at each point; the flexural rigidity of the beam is EI. Initially, when a load [latex]W_{1}=400 N[/latex] was applied at the point 1 the displacements were [latex]\Delta_{1}=0.75 mm , \Delta_{2}=-0.25 mm[/latex], and [latex]\Delta_{3}=-1.25 mm[/latex]. Find

(i) [latex]\Delta_{1}[/latex] given that [latex]\Delta_{2}=1 mm , \Delta_{3}=0.875 mm[/latex] for [latex]W_{2}=300 N[/latex].

(ii) [latex]\Delta_{1}[/latex] and [latex]\Delta_{2}[/latex] given that [latex]\Delta_{3}=2.5 mm[/latex] for [latex]W_{3}=500 N[/latex].

(iii) [latex]\Delta_{1}, \Delta_{2}[/latex], and [latex]\Delta_{3}[/latex] when [latex]W_{1}=W_{2}=W_{3}=600 N[/latex] are applied simultaneously.
Ans. (i) -0.1875 mm, (ii) -1.5625 mm, 1.4583 mm, (iii) -1.125 mm, 3.375 mm, 2.875 mm.

Accepted Answer

From Section 15.4

[latex]\Delta_{1}[/latex]=[latex]a_{11} W_{1}+a_{12} W_{2}+a_{13} W_{3}[/latex]

[latex]\Delta_{2}=a_{21} W_{1}+a_{22} W_{2}+a_{23} W_{3}[/latex] (i)

[latex]\Delta_{3}=a_{31} W_{1}+a_{32} W_{2}+a_{33} W_{3}[/latex]

For the initial condition Eqs (i) reduce to

[latex]0.75=a_{11} 400, \quad-0.25=a_{21} 400, \quad-1.25=a_{31} 400[/latex]

so that

[latex]a_{11}=0.75 / 400, \quad a_{21}=-0.25 / 400=a_{12}, \quad a_{31}=-1.25 / 400=a_{13}[/latex]

Case (i)

[latex]\Delta_{1}=a_{12} 300=(-0.25 / 400) imes 300=-0.1875[/latex]

Also from Eqs (i)

[latex]\Delta_{2}=1=a_{22} 300 ext { so that } a_{22}=1 / 300[/latex]

and

[latex]\Delta_{3}=0.875=a_{32} 300 ext { so that } a_{32}=0.875 / 300[/latex]

Case (ii)
From Eqs (i)

[latex]\Delta_{1}=(-1.25 / 400) imes 500=-1.5625[/latex]

[latex]\Delta_{2}=(0.875 / 300) imes 500=1.4583[/latex]

and

[latex]2.5=a_{33} 500 ext { so that } a_{33}=2.5 / 500[/latex]

Case (iii)

[latex]\Delta_{1}=(0.75 / 400) imes 600-(0.25 / 400) imes 600-(1.25 / 400) imes 600=-1.125 mm[/latex]

[latex]\Delta_{2}=(-0.25 / 400) imes 600+(1 / 300) imes 600+(0.875 / 300) imes 600=3.375 mm[/latex]

[latex]\Delta_{3}=(-1.25 / 400) imes 600+(0.875 / 300) imes 600+(2.5 / 500) imes 600=2.875 mm[/latex]

Question 15.P.26: The displacements at three points, 1, 2, and 3 on a beam wer...

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