Question 6.19: Use the fundamental equation of shear flow analysis for beam......

Plan the Solution We can follow the basic procedure that was followed in deriving Eq. 6.96, that is, we can derive an expression for e, like Eq. 6.89, based on resultant moments about O. This will involve the flange shear force. Next we can determine an expression for the flange shear flow, and then use the flange shear flow to determine the flange shear force needed in the expression for e. The shear center should lie “out-side” the cross section.

e = \frac{b^2t_f h_w \cos Φ} {I} \left(\frac{h_w}{4} + \frac{b \sin Φ}{3}\right)        (6.96)

e =\frac{V_1h_w \cos Φ}{V}            (6.89)

Neglecting a small flange contribution \left(\frac{1}{12}bt^3_f\right), we get the following value for the moment of inertia:

I = \frac{(4  mm)(250  mm)^3}{12} + 2(40 mm)(7 mm)(125 mm)²

I = 13.96(10^6)  mm^4 = 13.96(10^{-6})  m^4

From resultant moments about point O, through which V_w passes (Fig. 2),

so

e = \frac{V_f h_w} {V}          (1)

From Eq. 6.65,

q = \frac{VQ} {I}        Shear-Flow Formula        (6.65)

q_f = \frac{VQ_f} {I}        (2)

with Q_f based on the flange area A′ in Fig. 3, that is,

Q_f = A^′\bar{y}^′ = (7 mm)(s)(125 mm)

Q_f = 875s mm³

So,

(q_f)_{max} = (q_f)_{s=40  mm}       (3)

= 2.51(10^{-3}) V N/m

Since the shear flow varies linearly, as illustrated in Fig. 4, the flange shear force V_f is

V_f = \frac{1}{2}[2.507(10^{-3}) V N/mm](40 mm)

V_f = 0.0501 V N    (4)

Combining Eqs. (1) and (4) we get

e = 0.0501h_w = 12.5 mm

Review the Solution The eccentricity e should be positive so that the shear center lies outside the cross section. Our answer has the correct sign, and the magnitude seems reasonable in comparison with the dimensions of the beam. Many commercial channel sections have an e/h_w ratio in the 0.05–0.15 range.