Question 7.10: Determine the location of the flexural center for the two se...

Objective    Locate the shear center, Q, for the two shapes.

Given           Shapes in Figure 7–29; channel in 7–29(a); hat section in 7–29(b)

Analysis      The general location of the shear center for each shape is shown in Figure 7–27 along with the means of computing the value of e that locates Q relative to specified features of the shapes.

Results        Channel section (a): From Figure 7–27, the distance e to the flexural center is

e = \frac{b^{2}h^{2}t}{4I_{x}}

Note that the dimensions b and h are measured to the middle of the flange or web. Then, b = 40 mm and h = 50 mm. Because of symmetry about the x-axis, Ix can be found by the difference between the value of I for the large outside rectangle (54 mm × 42 mm) and the smaller rectangle removed (46 mm × 38 mm):

I_{x} = \frac{(42)(54)^{3} }{12} – \frac{(38)(46)^{3}}{12} = 0.243 \times 10^{6}   mm^{4}

Then,

e = \frac{(40)^{2}(50)^{2}(4) }{4(0.243 \times 10^{6})}  mm = 16.5 mm

This dimension is drawn to scale in Figure 7–29(a).

Hat section (b): Here, the distance e is a function of the ratios c/h and b/h:

\frac{c}{h} = \frac{10}{30} = 0.3 

\frac{b}{h} = \frac{30}{30} = 1.0

Then, from Figure 7–27, e/h = 0.45. Solving for e yields

e = 0.45h = 0.45 (30  mm) = 13.5 mm

This dimension is drawn to scale in Figure 7–29(b).

Comment    Now, can you devise a design for using either section as a beam and provide for the applica-tion of the load through the flexural center Q to produce pure bending?