Question 6.15: (a) Determine the shear stress distribution in the flanges a......
(a) Determine the shear stress distribution in the flanges and the web of a W14×26 beam^{30} subjected to the shear force V = -28 kips, as shown in Fig. 1.^{31} (b) Determine the percent of the vertical shear that is carried by the web of this cross section.
Plan the Solution Although it would be possible to solve Part (a) by using the formulas in Eqs. 6.74 and 6.78, it will be more instructive to use the basic shear-flow formulas, Eqs. 6.69 and 6.75, and the average shear-stress formula, Eq. 6.68.We can determine the total web shear force by integrating the shear flow, as indicated in Eq. 6.82. Since the shear on the cross section is negative (i.e., it points in +y direction in Fig. 1) the shear flows and shear stresses will act in the opposite sense to those shown in Fig. 6.54.
τ_f = \frac{q_f} {t_f} =\frac{3} {2} \left(\frac{Vs} {t_f}\right) \frac{h^2 – h^2 _w} {bh^3 – bh^3_w + t_wh^3} (6.74)
τ_w = \frac{3V} {2t_w} \left(\frac{bh^2 – bh^2_w – t_wh^2_w – 4t_wy^2} {bh^3 – bh^3_w + t_wh^3_w}\right) (6.78)
q_f = \frac{VQ_f} {I} (6.69)
q_w = \frac{VQ_w} {I} (6.75)
τ(x, s) = \frac{q(x, s)} {t(x, s)} (6.68)
V_w = \int^{hw/2} _{-hw/2} {q_w(y) dy} (6.82)
(a) Determine the shear flow and the shear stress in the flanges and in the web.
Flange Shear Flow and Shear Stress: The flange shear flow is given by Eq. 6.69:
q_f = \frac{VQ_f} {I} = \frac{VA^′_f \bar{y}_f^′} {I} = \frac{VA_1\bar{y}_1} {I} (1)
q_f = \frac{(-28 kips)(0.420 s in^2)(6.745 in.)} {245 in^4} = -0.324 s kips/in. (2)
τ_f = \frac{q_f} {t_f} = \frac{-0.324 s kips/in.} {0.420 in.} = -0.771 s ksi (3)
(τ_f)_{max} = |(τ_f)_{s=2.385 in.}| = 1.84 ksi (4)
τ_f = -0.771 s ksi, (τ_f)_{max} = 1.84 ksi (a)
Web Shear Flow and Web Shear Stress: The web shear flow is given by Eq. 6.75:
q_w = \frac{VQ_w} {I} = \frac{VA^′_w \bar{y}_w^′} {I} = \frac{V}{I}(A_2\bar{y}_2 + A_3\bar{y}_3) (5)
where A_2 is the area of the entire top flange. Therefore,
q_w = \left(\frac{-28 kips} {245 in^4}\right) [ (5.025 in.)(0.420 in.)(6.745 in.)
+ (6.535 in. – y)(0.255 in.)(0.5)(6.535 in. + y)]
or
q_w = \frac{(-28)[14.235 + 0.1275(42.71 – y^2)]} {245}= -(2.249 – 0.0146y²) kips/in.
\left\{\begin{matrix} τ_w = \frac{q_w}{t_w} = -(8.82 – 0.0571y^2) ksi\\ (τ_w)_{max} = |(τ_w)_{y=0}| = 8.82 ksi \\(τ_w)_{min} = |(τ_w)_{y=6.535 in.}| = 6.38 ksi \end{matrix} \right\} (a) \begin{matrix} (6) \\ (7a) \\(7b)\end{matrix}The results of the flange shear analysis and web shear analysis are summarized on Fig. 2. The negative shear (V = -28 kips) gives shear stresses in the opposite direction to those in Fig. 6.54.
(b) Determine the percent of vertical shear that is carried by the web. Let us use Eq. 6.82.
V_w = \int^{hw/2} _{-hw/2} {q_w(y) dy} = -\int^{6.535}_{-6.535} {(2.249 – 0.0146y^2) dy}= -2.249(6.535)(2) + \frac{2}{3} (0.0146)(6.535)³
= -29.40 + 2.71 = -26.69 kips
Comparing V_w with V, we see that 95% of the vertical shear is carried by the web of this beam.
Review the Solution We can spot-check the above results by using Eq. 6.80 to compute (τ_w)_{max}. When we do so, we get (τ_w)_{max} = 9.02 ksi, which does not agree very well with the value of 8.82 ksi that we got in Eq. (7a).
So let us compare the value of I = 245 in^4 from Appendix D.1 with the value calculated from Eq. 6.70. For the latter we get I = 240 in^4. The difference in I values accounts for the difference in (τ_w)_{max} values. This difference stems from the fact that the actual cross section has web-to-flange fillets that provide a slight increase in the value of the moment of inertia. This was not accounted for in Eq. 6.70.
(τ_w)_{max} = \frac{3V} {2t_w} \left(\frac{bh^2 – bh^2_w + t_wh^2_w} {bh^3 – bh^3_w + t_wh^3_w}\right) (6.80)
I = \frac{1} {12} (bh^3 – bh^3_w + t_wh^3_w) (6.70)
