(a) A rectangular-section steel beam, 50 mm wide by 20 mm deep, is used as a simply supported beam over a span of 2 m with the 20 mm dimension vertical. Determine the value of the central concentrated load which will produce initiation of yield at the outer fibres of the beam. (b) If the central

Question

(a) A rectangular-section steel beam, 50 mm wide by 20 mm deep, is used as a simply supported beam over a span of 2 m with the 20 mm dimension vertical. Determine the value of the central concentrated load which will produce initiation of yield at the outer fibres of the beam.

(b) If the central load is then increased by 10% find the depth to which yielding will take place at the centre of the beam span.
(c) Over what length of beam will yielding then have taken place?
(d) What are the maximum deflections for each load case?
For steel gv in simple tension and compression [latex]= 225 MN/m^{2}[/latex] and [latex]E = 206.8 GN/m^{2}[/latex].

Accepted Answer

(a) From eqn. (3.1) the B.M. required to initiate yielding is

[latex]M_{E}=\frac{BD^{2} }{6} \sigma _{y}[/latex] (3.1)

[latex]\frac{BD^{2} }{6} \sigma _{y} =\frac{50 imes20^{2} imes 10^{-9} }{6} imes 225 imes 10^{6} =750 Nm[/latex]

But the maximum B.M. on a beam with a central point load is WL/4, at the centre.

[latex]\frac{W imes 2}{4} =750[/latex]
i.e. [latex]W=1500N[/latex]

The load required to initiate yielding is 1500 N.
(b) If the load is increased by 10% the new load is

[latex]W^{′} =1500+150=1650 N[/latex]

The maximum B .M. is therefore increased to

[latex]M^{′} =\frac{W^{′} L}{4} =\frac{1650 imes 2}{4} =825 Nm[/latex]

and this is sufficient to produce yielding to a depth d , and from eqn. (3.2),

[latex]M_{pp} =\frac{B\sigma _{y} }{12} \left[3D^{2} -d^{2} ight] [/latex] (3.2)

[latex]M_{pp} =\frac{B\sigma _{y} }{12} \left[3D^{2} -d^{2} ight]=825 N m[/latex]
[latex]825=\frac{50 imes 10^{-3} imes 225 imes 10^{6} }{12} \left[3 imes 2^{2} -d^{2} ight] 10^{-4}[/latex]

where d is the depth of the elastic core in centimetres,

[latex]8.8=12-d^{2}[/latex]
[latex]d^{2}=3.2 ext{ and } d=1.79 cm[/latex]

[latex]depth of yielding =\frac{1}{2} \left(D-d ight) =\frac{1}{2}\left(20-17.9 ight) =1.05 mm[/latex]

(c) With the central load at 1650 N the yielding will have spread from the centre as shown in Fig. 3.37. At the extremity of the yielded region, a distance x from each end of the beam, the section will just have yielded at the extreme surface fibres, i.e. the moment camed at this section will be the maximum elastic moment and given by eqn. (3.1) - see part (a) above.

Now the B.M. at the distance x from the support is

[latex]\frac{1650 x}{2} =\frac{BD^{2} }{6} \sigma _{y}=750[/latex]
[latex]x=\frac{2 imes 750}{1650}=0.91 m[/latex]

Therefore length of beam over which yielding has occurred

[latex]=2-2 imes 0.91=0.18 m=180 mm[/latex]

(d) For W = 1500 N the beam is completely elastic and the maximum deflection, at the centre, is given by the standard form of eqn. (5.15)[latex]^{\dagger }[/latex]:

[latex]\delta =\frac{W L^{3} }{48 EI} =\frac{1500 imes 2^{3} imes 12 }{48 imes 206.8 imes 10^{9} imes 50 imes 20^{3} imes 10^{-12} }[/latex]
[latex]=0.0363 m=36.3 mm[/latex]

With W = 1650 N and the beam partially plastic, deflections are calculated on the basis of the elastic core only

[latex]\delta =\frac{W^{′} L^{3} }{48 EI^{′}}=\frac{1650 imes 2^{3} imes 12}{48 imes 206.8 imes 10^{9} imes 50 imes 17.9^{3} imes 10^{-12}}[/latex]
[latex]=0.0556 m=55.6 mm[/latex]

[latex] ^{\dagger } [/latex] E.J. Hem, Mechanics of Materials I, Butterworth-Heinemann, 1997.

Question 3.1: (a) A rectangular-section steel beam, 50 mm wide by 20 mm de...

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