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Question 3.2: (a) Determine the “shape factor” of a T-section beam of dime...

(a) Determine the “shape factor” of a T-section beam of dimensions 100 mm × 150 mm × 12 mm as shown in Fig. 3.38.
(b) A cantilever is to be constructed from a beam with the above section and is designed to carry a uniformly distributed load over its complete length of 2 m. Determine the maximum u.d.1. that the cantilever can carry if yielding is permitted over the lower part of the web to a depth of 25 mm. The yield stress of the material of the cantilever is 225 MN/m².

3.38
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(a)                            Shape factor = fully plastic moment  maximum elastic moment \text { Shape factor }=\frac{\text { fully plastic moment }}{\text { maximum elastic moment }}

To determine the maximum moment carried by the beam while completely elastic we must first determine the position of the N.A. Take moments of area about the top edge (see Fig. 3.38):

(100×12×6)+(138×12×81)=[(100×12)+(138×12)]yˉ(100 \times 12 \times 6) +(138\times 12\times 81)=[(100\times 12)+(138\times 12)] \bar{y}
7200+134136=(1200+1656)yˉ7200+134136=(1200+1656)\bar{y}
∴       yˉ=49.5  mm\bar{y} =49.5   mm
∴       INA=[100×49.533+12×100.533−88×37.533]10−12m4I_{NA} =[\frac{100\times 49.5^{3}}{3}+\frac{12\times 100.5^{3}}{3}-\frac{88\times 37.5^{3}}{3} ]10^{-12}m^{4}

=13[121.29+121.81−46.4]10−7=\frac{1}{3}[121.29+121.81-46.4] 10^{-7}
=6.56×10−6m4=6.56\times 10^{-6} m^{4}

Now from the simple bending theory the moment required to produce the yield stress at the edge of the section (in this case the lower edge), i.e. the maximum elastic moment, is

ME=σIymax=σy×6.56×10−6100.5×10−3=0.065×10−3σyM_{E} =\frac{\sigma I}{y_{max} } =\sigma _{y}\times \frac{6.56\times 10^{-6}}{100.5\times 10^{-3}} =0.065\times 10^{-3} \sigma _{y}

When the section becomes fully plastic the N.A. is positioned such that

area below N.A. = half total area

i.e. if the plastic N.A. is a distance yˉp\bar{y}_{p} above the base, then

ypˉ×12=12(1200+1656)\bar{y_{p} } \times 12=\frac{1}{2}(1200+1656)
∴              ypˉ=119  mm\bar{y_{p} }=119   mm

The fully plastic moment is then obtained by considering the moments of forces on convenient rectangular parts of the section, each being subjected to a uniform stress σy \sigma _{y},

i.e.            MFP=[σy(100×12)(31−6)+σy(31−12)×12×(31−12)2+σy(119×12)1192]10−9 M_{FP} =[\sigma _{y}(100\times 12)(31-6)+\sigma _{y} (31-12) \times 12\times \frac{(31-12)}{2} +\sigma _{y}(119\times 12)\frac{119}{2} ]10^{-9}
=σy(30000+2166+84966)10−9 =\sigma _{y}(30000+2166+84966) 10^{-9}
=0.117+10−3σy =0.117+10^{-3}\sigma _{y}
∴              shape factor =MFPME=0.117×10−30.065×10−3=1.8 =\frac{M_{FP} }{M_{E} } =\frac{0.117\times 10^{-3}}{0.065\times 10^{-3}} =1.8

(b) For this part of the question the load on the cantilever is such that yielding has progressed to a depth of 25 mm over the lower part of the web. It has been shown in § 3.4§  3.4 that whilst plastic penetration proceeds, the N.A. of the section moves and is always positioned by the rule:

compressive force above N.A. = tensile force below N.A.

Thus if the partially plastic N.A. is positioned a distance y above the extremity of the yielded area as shown in Fig. 3.39, the forces exerted on the various parts of the section may be established (proportions of the stress distribution diagram being used to determine the various values of stress noted in the figure).

Force on yielded area                                                  F1F_{1} = stress × area

 =225×106(12×25×10−6)=225\times 10^{6}(12\times 25\times 10^{-6})
=67.5  kN=67.5   kN

Force on elastic portion of web below N.A

  F2F_{2} = average stress × area

=225×1062(12×y×10−6)=\frac{225\times 10^{6}}{2} (12\times y\times 10^{-6})
=1.35y kN=1.35 y  kN

where y is in millimetres.
Force in web above N.A.

F3F_{3} = average stress × area

=(113−y)2y(225×106)(113−y)12×10−6=\frac{(113-y)}{2y}(225\times 10^{6}) (113-y)12\times 10^{-6}
=1.35(113−y)2ykN=1.35\frac{(113-y)^{2}}{y} kN

Force in flange

F4= average stress × area =12[(113−y)y+(125−y)y](225×106)100×12×10−6 approximately =(238−2y)2y225×106×100×12×10−6=135(238−2y)y kN\begin{aligned}F_{4} &=\text { average stress } \times \text { area } \\&=\frac{1}{2}\left[\frac{(113-y)}{y}+\frac{(125-y)}{y}\right]\left(225 \times 10^{6}\right) 100 \times 12 \times 10^{-6} \text { approximately } \\&=\frac{(238-2 y)}{2 y} 225 \times 10^{6} \times 100 \times 12 \times 10^{-6} \\&=135 \frac{(238-2 y)}{y}  kN\end{aligned}

Now for the resultant force across the section to be zero,

F1+F2=F3+F467.5+1.35y=1.35(113−y)2y+135(238−2y)y∴67.5y+1.35y2=17.24×103−305y+1.35y2+32.13×103−270y642.5y=49370y=76.8  mm\begin{aligned}F_{1}+F_{2} &=F_{3}+F_{4} \\67.5+1.35 y &=\frac{1.35(113-y)^{2}}{y}+\frac{135(238-2 y)}{y} \\\therefore \quad 67.5 y+1.35 y^{2} &=17.24 \times 10^{3}-305 y+1.35 y^{2}+32.13 \times 10^{3}-270 y \\642.5 y &=49370 \\ y &=76.8   mm\end{aligned}

Substituting back,

F1=67.5 kNF2=103.7 kNF3=23 kNF4=148.1 kN\begin{array}{ll}F_{1}=67.5  kN & F_{2}=103.7  kN \\F_{3}=23  kN & F_{4}=148.1  kN\end{array}

The moment of resistance of the beam can now be obtained by taking the moments of these forces about the N.A. Here, for ease of calculation, it is assumed that F4F_{4} acts at the mid-point of the web. This, in most cases, is sufficiently accurate for practical purposes.

 Moment of resistance ={F1(y+12.5)+F2(2y3)+F3[23(113−y)]+F4[(113−y)+6]}10−3kNm=(6030+5312+554+6243)10−3kNm=18.14  kNm\begin{aligned}\text { Moment of resistance }=&\left\{F_{1}(y+12.5)+F_{2}\left(\frac{2 y}{3}\right)+F_{3}\left[\frac{2}{3}(113-y)\right]\right.\\&\left.+F_{4}[(113-y)+6]\right\} 10^{-3} kNm \\=&(6030+5312+554+6243) 10^{-3} kNm \\=& 18.14   kNm\end{aligned}

Now the maximum B.M. present on a cantilever carrying a u.d.1. is wL2/2wL^{2}/2 at the support

∴                    wL22=18.15×103\frac{w L^{2}}{2}=18.15 \times 10^{3}

The maximum u.d.1. which can be carried by the cantilever is then

w=18.15×103×24=9.1 kN/mw=\frac{18.15 \times 10^{3} \times 2}{4}=9.1  kN / m
3.39

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