Question 6.20: (a) Prove that the shear center of an equal-leg angle (Fig. ......
(a) Prove that the shear center of an equal-leg angle (Fig. 1) is at the corner of the angle, and prove that the resultant of the shear flow in the legs of the angle is equal to the total shear force on the section. The shear force acts normal to the axis of symmetry of the cross section. Assume that t \ll b. (b) Determine the maximum shear stress in the cross section in Fig. 1.
Plan the Solution There is a resultant shear force along each leg of the angle. Since the lines of action of these two forces pass through the corner O, this point must be the shear center. We could develop an expression for this using basic shear-flow concepts (i.e., Eq. 6.65), or we can make use of Eq. 6.95, setting h_w = 0 and Φ = 45°. The maximum shear stress should occur at the neutral axis.
q = \frac{VQ} {I} Shear-Flow Formula (6.65)
V_1 = \int^b_0{q_1(s) ds} = \frac{Vt_f b^2}{I} \left(\frac{h_w}{4} + \frac{b \sin Φ} {3}\right) (6.95)
(a) Determine shear-center location and prove that the resultant of flange shears is V. Since V_1 and V_2 both pass through point O, as indicated in Fig. 2, the resultant, V, must pass through this point.
Therefore, the shear center lies at the intersection of the legs of the angle.^{35} The flange shear is given by Eq. 6.95, with h_w = 0.
V_f = \frac{Vtb^3 \sin 45°}{3I} (1)
where the moment of inertia is obtained by specializing Eq. 6.92 to give
I = \frac{t_wh^3_w} {12} + 2bt_f \left[\left(\frac{h_w}{2}\right)^2 + \left(\frac{bh_w}{2}\right) \sin Φ + \left(\frac{b^2}{3}\right) \sin^2 Φ\right] (6.92)
I = \frac{2}{3} b^3t \sin^2 45° = \frac{b^3t}{3} (2)
Combining Eqs. (1) and (2) we get
V_f = \frac{V} {\sqrt{2}} (3)
Since
+↓ ∑F_y = 2V_f \sin 45° = Vwe have shown that combining Eqs. 6.92 and 6.95 leads to the correct resultant shear force.
(b) Determine the maximum shear stress. The shear flow distribution is given by Eq. 6.94, which specializes to
q_1(s) = \frac{VQ_1(s)}{I}= \frac{Vt_f s}{I} \left[\frac{h_w}{2}+ \left(b – \frac{s}{2}\right) \sin Φ\right] (6.94)
q_f (s) = \frac{Vts} {I} \left(b – \frac{s}{2}\right) sin 45° (4)
where I is given by Eq. (2). This shear-flow distribution is depicted in Fig. 3. Since τ_f = q_f /t,
(τ_f)_{max} = \frac{(q_f)_{s=b}}{t}or
(τ_f)_{max} = \left(\frac{3\sqrt{2}}{4}\right) \frac{V}{bt} (5)
Review the Solution As a rough check on our answer for τ_{max}, we can compare the preceding result, Eq. (5), with τ_{max} for a rectangle. From Example Problem 6.14,
(τ_{max})_{\text{rectangle}} = \frac{3}{2}\frac{V}{A}Since the area of the equal-leg angle is 2bt, Eq. (5) can be expressed as
(τ_{f max})_{\text{angle}} = \frac{3\sqrt{2}} {2} \frac{V}{A}Since the shear flows in the angle are not parallel with the resultant V, it is reasonable that the shear stress in the angle section is greater than that in a rectangle by a factor of \sqrt{2}.
See Homework Problem 6.12–16 for a chance to use the basic shear flow formula, Eq. 6.65, to obtain the above results.
