In Example 6.3, the width-to-height ratio in the cross-section of the beam is b/h = 5/1 = 5. This may be considered as a narrow-beam cross-section. If the b/h ratio were large, the cross-section may be considered to be wide beam. What are the results of Example 6.3 if one considers the beam to be a

Question

In Example 6.3, the width-to-height ratio in the cross-section of the beam is b/h = 5/1 = 5. This may be considered as a narrow-beam cross-section. If the b/h ratio were large, the cross-section may be considered to be wide beam. What are the results of Example 6.3 if one considers the beam to be a wide beam?

Accepted Answer

In the case of the wide beams, we consider

[latex]\kappa_y=0[/latex].

Then, from Equation (6.24),

[latex] \left[\begin{array}{c} \epsilon_{x}^{0} \ \epsilon_{y}^{0} \ \gamma_{x y}^{0} \ \kappa_{x} \ 0 \ \kappa_{x y} \end{array} ight]=\left[\begin{array}{cccccc} J_{11} & J_{12} & J_{13} & J_{14} & J_{15} & J_{16} \ J_{21} & J_{22} & J_{23} & J_{24} & J_{25} & J_{26} \ J_{31} & J_{32} & J_{33} & J_{34} & J_{35} & J_{36} \ J_{41} & J_{42} & J_{43} & J_{44} & J_{45} & J_{46} \ J_{51} & J_{52} & J_{53} & J_{54} & J_{55} & J_{56} \ J_{61} & J_{62} & J_{63} & J_{64} & J_{65} & J_{66} \end{array} ight]\left[\begin{array}{c} 0 \ 0 \ 0 \ M_{x} \ M_{y} \ 0 \end{array} ight], [/latex] (6.35)

we get

[latex] \epsilon_{x}^{0}=J_{14} M_{x}+J_{15} M_{y} [/latex] (6.36a)

[latex] \epsilon_{y}^{0}=J_{24} M_{x}+J_{25} M_{y} [/latex] (6.36b)

[latex] \gamma_{x y}^{0}=J_{34} M_{x}+J_{35} M_{y} [/latex] (6.36c)

[latex] \kappa_{x}=J_{44} M_{x}+J_{45} M_{y} [/latex] (6.36d)

[latex] 0=J_{54} M_{x}+J_{55} M_{y} [/latex] (6.36e)

[latex] \kappa_{x y}=J_{64} M_{x}+J_{65} M_{y} . [/latex] (6.36f)

To find the neutral axis, [latex]∈_x = 0[/latex], we use Equation (6.36a) and Equation (6.36e) to give

[latex] z_{n}=-\frac{J_{14} J_{55}-J_{15} J_{54}}{J_{44} J_{55}-J_{45} J_{54}} [/latex] (6.37)

[latex] M_{ ext {beam }}=b M_{x}=b \frac{J_{55}}{J_{44} J_{55}-J_{54} J_{45}} \kappa_{x} [/latex] (6.38)

From Equation (6.9a), Equation (6.11), and Equation (6.38),

[latex] E_{x}=\frac{12}{h^{3}} \frac{J_{55}}{\left(J_{44} J_{55}-J_{45} J_{54} ight)} \ \space \=\frac{12}{(0.001)^{3}} \frac{2.551 imes 10^{-1}}{\left(1.194 imes 10^{-1} ight)\left(2.551 imes 10^{-1} ight)-\left(-4.355 imes 10^{-2} ight)\left(-4.355 imes 10^{-2} ight)} \ \space \ =1.071 imes 10^{11} Pa. [/latex]

Thus, from Equation (6.20), we get

[latex] \delta=\frac{(5)(200)(0.1)^{4}}{(384)\left(1.072 imes 10^{11} ight)\left(4.167 imes 10^{-13} ight)} \ =5.830 imes 10^{-3} m \ = 5.830 mm .[/latex]

From Example 6.3, the maximum bendings’ moment per unit width is

[latex] M_{x \mid \max }=50 \frac{N-m}{m} [/latex] .

From Equation (6.36e),

[latex] \left.M_{y} ight|_{\max }=-\frac{J_{54}}{J_{55}} M_{x} \ \space \ =-\frac{-4.355 imes 10^{-2}}{2.551 imes 10^{-1}}(50) \ \space \ =8.497 \frac{N-m}{m}[/latex].

From Equation (6.35),

[latex] \left[\begin{array}{c} \epsilon_{x}^{0} \ \epsilon_{y}^{0} \ \gamma_{x y}^{0} \ \kappa_{x} \ 0 \ \kappa_{y} \end{array} ight]= \ \space \ \left[\begin{array}{cccccc} 1.068 imes 10^{-8} & -3.409 imes 10^{-9} & 7.009 imes 10^{-10} & 4.298 imes 10^{-6} & -7.241 imes 10^{-6} & -9.809 imes 10^{-6} \ -3.409 imes 10^{-9} & 1.829 imes 10^{-8} & 4.042 imes 10^{-10} & -6.097 imes 10^{-6} & 1.142 imes 10^{-5} & -3.083 imes 10^{-6} \ 7.009 imes 10^{-10} & 4.042 imes 10^{-10} & 5.035 imes 10^{-8} & -6.339 imes 10^{-6} & -3.460 imes 10^{-6} & -4.989 imes 10^{-5} \ 4.298 imes 10^{-6} & -6.097 imes 10^{-6} & -6.339 imes 10^{-6} & 1.194 imes 10^{-1} & -4.335 imes 10^{-2} & -1.940 imes 10^{-2} \ -7.241 imes 10^{-6} & 1.142 imes 10^{-5} & -3.460 imes 10^{-6} & -4.355 imes 10^{-2} & 2.551 imes 10^{-1} & -5.480 imes 10^{-3} \ -9.809 imes 10^{-6} & -3.083 imes 10^{-6} & -4.989 imes 10^{-5} & -1.940 imes 10^{-2} & -5.480 imes 10^{-3} & 6.123 imes 10^{-1} \end{array} ight]\left[\begin{array}{c} 0 \ 0 \ 0 \ 50 \ 8.497 \ 0 \end{array} ight] \ \space \ =\left[\begin{array}{c} 1.534 imes 10^{-4} \ -2.078 imes 10^{-4} \ -3.463 imes 10^{-4} \ 5.602 \ 0 \ -1.017 \end{array} ight] [/latex]

The global strains (Equation 6.15) at the top of the third ply (–30°) are

[latex] \left[\begin{array}{l} \epsilon_{x} \ \epsilon_{y} \ \gamma_{x y} \end{array} ight]=\left[\begin{array}{c} \epsilon_{x}^{0} \ \epsilon_{y}^{0} \ \gamma_{x y}^{0} \end{array} ight]+ z \left[\begin{array}{c} \kappa _{x} \ \kappa_{y} \ \kappa_{x y} \end{array} ight] \ \space \ =\left[\begin{array}{c} 1.534 imes 10^{-4} \ -2.078 imes 10^{-4} \ -3.463 imes 10^{-4} \end{array} ight]+(-0.00025)\left[\begin{array}{c} 5.602 \ 0 \ -1.017 \end{array} ight] \ \space \ =\left[\begin{array}{c} -1.247 imes 10^{-3} \ -2.078 imes 10^{-4} \ -9.221 imes 10^{-5} \end{array} ight] \frac{m}{m} [/latex].

The global stresses (Equation 6.18) at the top of the third ply (–30°) are

[latex] \left[\begin{array}{l} \sigma_{x} \ \sigma_{y} \ au_{x y} \end{array} ight]=[\bar{Q}]\left[\begin{array}{l} \epsilon_{x} \ \epsilon_{y} \ \gamma_{x y} \end{array} ight] \ \space \ =\left[\begin{array}{ccc} 1.094 imes 10^{11} & 3.246 imes 10^{10} & -5.419 imes 10^{10} \ 3.246 imes 10^{10} & 2.365 imes 10^{10} & -2.005 imes 10^{10} \ -5.419 imes 10^{10} & -2.005 imes 10^{10} & 3.674 imes 10^{10} \end{array} ight]\left[\begin{array}{l} -1.247 imes 10^{-3} \ -2.078 imes 10^{-4} \ -9.221 imes 10^{-3} \end{array} ight] \ \space \ =\left[\begin{array}{c} -1.382 imes 10^{8} \ -4.354 imes 10^{7} \ 6.833 imes 10^{7} \end{array} ight] [/latex].

The relative differences in the stresses obtained using wide and narrow beam assumptions are

[latex] \left|\in_{a} ight|_{\sigma_{x}}=\left|\frac{\sigma_{x \mid ext { narrow }}-\sigma_{x \mid ext { wide }}}{\sigma_{x \mid ext { narrow }}} ight| imes 100 \ \space \ =\left|\frac{-1.280 imes 10^{8}-\left(-1.382 imes 10^{8} ight)}{-1.280 imes 10^{8}} ight| imes 100 \ \space \ =7.97\% \ \space \ \left|\epsilon_{a} ight|_{\sigma_{x}}=\left|\frac{\sigma_{y \mid ext { narrow }}-\sigma_{y \mid ext { wide }}}{\sigma_{y \mid ext { narrow }}} ight| imes 100\ \space \=\left|\frac{-3.431 imes 10^{7}-\left(-4.354 imes 10^{7} ight)}{-3.431 imes 10^{7}} ight| imes 100 \ \space \ =26.90\% \ \space \ \left|\epsilon_{a} ight|_{ au_{x y}}=\left|\frac{ au_{x y \mid ext { narrow }}- au_{x y \mid w i d e}}{ au_{x y \mid ext { narrow }}} ight| imes 100 \ \space \ =\left|\frac{6.170 imes 10^{7}-6.836 imes 10^{7}}{6.170 imes 10^{7}} ight| imes 100 \ \space \ =10.79\%[/latex].
______________________________

(6.24): [latex] \left[\begin{array}{c} \epsilon_{x}^{0} \ \epsilon_{y}^{0} \ \gamma_{x y}^{0} \ \kappa_{x} \ \kappa_{y} \ \kappa_{x y} \end{array} ight]=\left[\begin{array}{cccccc} J_{11} & J_{12} & J_{13} & J_{14} & J_{15} & J_{16} \ J_{21} & J_{22} & J_{23} & J_{24} & J_{25} & J_{26} \ J_{31} & J_{32} & J_{33} & J_{34} & J_{35} & J_{36} \ J_{41} & J_{42} & J_{43} & J_{44} & J_{45} & J_{46} \ J_{51} & J_{52} & J_{53} & J_{54} & J_{55} & J_{56} \ J_{61} & J_{62} & J_{63} & J_{64} & J_{65} & J_{66} \end{array} ight]\left[\begin{array}{c} N_{x} \ N_{y} \ N_{x y} \ M_{x} \ M_{y} \ M_{x y} \end{array} ight] . [/latex]

(6.9):
[latex] \kappa_{x}=-\frac{\partial^{2} w_{0}}{\partial x^{2}} \ \kappa_{y}=-\frac{\partial^{2} w_{0}}{\partial y^{2}} \ \kappa_{x y}=-2 \frac{\partial^{2} w_{0}}{\partial x \partial y} [/latex]

(6.11): [latex] \frac{d^{2} w_{0}}{d x^{2}}=-\frac{M_{x} b}{E_{x} I} [/latex] where
b = width of beam
[latex]E_x[/latex] = effective bending modulus of beam
I = second moment of area with respect to the x–y-plane

(6.20): [latex]\delta=\frac{5 q L^{4}}{384 E_{x} I} [/latex]

(6.15):
[latex]∈_x = zκ_x \ ∈_y = zκ_y \ γ_{xy}=zκ_{xy}[/latex]

(6.18): [latex] \left[\begin{array}{c} \sigma_{x} \ \sigma_{y} \ au_{x y} \end{array} ight]_{k}=[T]^{-1}\left[\begin{array}{c} \sigma_{1} \ \sigma_{2} \ au_{12} \end{array} ight]_{k} [/latex]

Question 6.4: In Example 6.3, the width-to-height ratio in the cross-secti...

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