Question

Two metal pieces each of size 125 m $\times$ 50 mm are glued together to form an inverted T-beam. If the maximum allowable stress both in tension and compression for web material is 200 MPa and that for flange material is 150 MPa then the magnitude of maximum sagging moment in kN-m that can be applied safely on the beam considering young's modulus $E_{web}=2\times {10}^{5}MPa$ & $E_{flange}=0.8\times {10}^{5}MPa$ respectively is

• A. 51.04

Answer 1

The correct option is A 51.04

${\sigma }_{max1}=200MPa$
${\sigma }_{max2}=150MPa$
$E_1=2\times {10}^{5}MPa$
$E_2=0.8\times {10}^{5}MPa$
Let us transform material 1 into material '2'

Transformed width of web = $50\times \frac{2\times {10}^5}{0.8\times {10}^5}=125mm$
Hence, the transformed section is
Possibility of occurrence of maximum stresses is at location A or at B or at 'C'
$\therefore$ Stress at A in transformed section
${\sigma }_{AT}=\frac{M}{z}=\frac{M}{\frac{125\times {175}^2}{6}}$

Actual stress at A in original section,

${\sigma }_A={\sigma }_{AT}\times \frac{E_1}{E_2}=\frac{6M}{125\times {175}^2}\times \frac{2\times {10}^5}{0.8\times {10}^5}$

Also, given

${\sigma }_A\le 200MPa$

$\Rightarrow \frac{6M\times 2\times {10}^5}{125\times {175}^2\times 0.8\times {10}^5}\le 200$

$\Rightarrow M\le 51.04kN-m$

Similarly stress at 'B' in material -1

${\sigma }_B=\frac{M{y}_b}{I}\times \frac{E_1}{E_2}=\frac{M\times 37.5}{\frac{125\times {175}^3}{12}}\times \frac{2\times {10}^5}{0.8\times {10}^5}\le 200$

$\Rightarrow M\le 119.10kN-m$

Again stress at C in material - 2

${\sigma }_C=\frac{M}{z}=\frac{M}{\frac{125\times {175}^2}{6}}\le 150$

$M\le 95.70kN-m$

Therefore , maximum allowable sagging BM on the beam is min {51.04,119.10, 95.70 kN-m}

$M_{max}=51.04kN-m$