An extruded bar of aluminium alloy has the cross-section as shown in Figure 13.26. About 2.10 m length of this bar is used as a simple beam on a span of 2.0 m. A concentrated load 5.0 kN is applied at midlength of the span. Calculate the stresses (both maximum tensile and compressive) for the beam.

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Let us divide our calculation in two phases – in the first phase we shall calculate the central properties of area and in the second phase we shall perform our stress calculation.
1. Coordinates and area moment of inertia calculation
(a) Coordinates of the centroid of the section [latex] C (\bar{x}, \bar{y}) [/latex]: By composite area method [latex]{}^3[/latex], we obtain

[latex] \bar{z}=\frac{(70)(10)(35)+(50)(10)(5)+(60)(10)(30)+(20)(10)(55)}{(70)(10)+(50)(10)+(60)(10)+(20)(10)} mm [/latex]

= 28.0 mm

and [latex] \bar{y}=\frac{(70)(10)(5)+(50)(10)(35)+(60)(10)(65)+(20)(10)(50)}{(70)(10)+(50)(10)+(60)(10)+(20)(10)} mm [/latex]

= 35.0 mm

(b) Area moment of inertia: Now,

[latex] \begin{aligned} \bar{I}_{y y}=& {\left[\left\{\frac{1}{12}(10)(70)^3+(70)(10)(7)^2 ight\}+\left\{\frac{1}{12}(50)(10)^3+(50)(10)(23)^2 ight\} ight.} \ &\left.+\left\{\frac{1}{12}(10)(60)^3+(60)(10)(2)^2 ight\}+\left\{\frac{1}{12}(20)(10)^3+(20)(10)(27)^2 ight\} ight] \end{aligned} [/latex]

or [latex] \bar{I}_{y y}=0.919 imes 10^6 mm ^4 [/latex]

[latex] \begin{aligned} \bar{I}_{z z}=& {\left[\left\{\frac{1}{12}(70)(10)^3+(70)(10)(30)^2 ight\}+\left\{\frac{1}{12}(10)(50)^3+(10)(50)(0)^2 ight\} ight.} \ &\left.+\left\{\frac{1}{12}(60)(10)^3+(60)(10)(30)^2 ight\}+\left\{\frac{1}{12}(10)(20)^3+(20)(10)(15)^2 ight\} ight] \end{aligned} [/latex]

or [latex] \bar{I}_{z z}=1.337 imes 10^6 mm ^4 [/latex]

and [latex] \begin{aligned} \bar{I}_{y z}=& {[\{0+(70)(10)(7)(-30)\}+\{0+(10)(50)(-23)(0)\}} \ &+\{0+(60)(10)(2)(30)\}+\{0+(20)(10)(27)(15)\}] mm ^4 \end{aligned} [/latex]

or [latex] \bar{I}_{y z}=-0.03 imes 10^6 mm ^4 [/latex]

Therefore, we get for the section

[latex] \bar{I}_{y y}=0.919 imes 10^6 mm ^4, \quad \bar{I}_{z z}=1.337 imes 10^6 mm ^4, \quad \bar{I}_{y z}=-0.03 imes 10^6 mm ^4 [/latex]

2. Stress calculation

We first depict the loading situation on the beam in the following isometric sketch [Figure 13.27(a)]:

In Figure 13.27(b), we show the bending moments only on the section at midspan of the beam.
Clearly,

[latex] M_y=+\left(2.5 \cos 80.2^{\circ} ight)(1) kN m =0.426 imes 10^6 N mm [/latex]

and [latex] M_z=+\left(2.5 \sin 80.2^{\circ} ight)(1) kN m =2.464 imes 10^6 N mm [/latex]

Thus, flexural stress at any point (y, z) on the beam section is given by

[latex] \sigma_{x x}=\frac{1}{\bar{I}_{y y} M_{z z}-\bar{I}_{y z}^2}\left[\left(z \bar{I}_{z z}-y \bar{I}_{y z} ight) M_y+\left(z \bar{I}_{y z}-y \bar{I}_{y y} ight) M_z ight] [/latex]

For NA, [latex] \sigma_{x x}=0 [/latex] and its angular orientation with respect to z-axis is given by

[latex] an \phi=\frac{y}{z}=\frac{\bar{I}_{z z} M_y+\bar{I}_{y z} M_z}{\bar{I}_{y z} M_y+\bar{I}_{y y} M_z} [/latex]

[latex] =\frac{(1.337)(0.426)+(-0.03)(2.464)}{(-0.03)(0.426)+(0.919)(2.464)}=0.22 [/latex]

Thus, [latex] \phi=12.4^{\circ} [/latex]. The NA is shown in Figure 13.27(b) above. Clearly, points M (35,–28) mm and N (−35, 42) mm are critical points. So,

[latex] \sigma_{ M }=\frac{1}{(0.919)(1.337)-(0.03)^2}\left[\begin{array}{l} \{(-28)(1.337)-(35)(-0.03)\}(0.426) \ +\{(-28)(-0.03)-(35)(0.919)\}(2.464) \end{array} ight] [/latex]

or [latex] \sigma_{ M }=-75.49 MPa [/latex]

and [latex] \sigma_{ N }=\frac{1}{(0.919)(1.337)-(-0.03)^2}\left[\begin{array}{l} \{(42)(1.337)-(-35)(-0.03)\}(0.426) \ +\{(42)(-0.03)-(-35)(0.919)\}(2.464) \end{array} ight] [/latex]

or [latex] \sigma_{ N }=81.14 MPa [/latex]

The maximum compressive stress in the beam is 75.49 MPa and maximum tensile stress is 81.14 MPa.

[latex]{}^3[/latex] Refer to Reference (1) in the Bibliography section of the book for detailed discussion.

Question 13.10: An extruded bar of aluminium alloy has the cross-section as ...

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