Question 6.9: A thin-walled semicircular cross section of radius r and thi......

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1. Conceptualize: The shear center is located somewhere on the axis of symmetry (the z axis). To determine the exact position, assume that the beam is bent by a shear force V_y acting parallel to the y axis and producing bending about the z axis as the neutral axis (Fig. 6-45b).
2. Categorize:
Shear stresses: The first step is to determine the shear stresses \tau acting on the cross section (Fig. 6-45b). Consider a section bb defined by the distance s measured along the centerline of the cross section from point a. The central angle subtended between point a and section bb is denoted θ. Therefore, the distance s equals rθ, where r is the radius of the centerline and θ is measured in radians.
To evaluate the first moment of the cross-sectional area between point a and section bb, identify an element of area dA (shown shaded in the figure) and integrate as
Q_z = \int ydA = \int\limits_{0}^{θ}(r~ cos ɸ)(tr~ dɸ) = r^2 ~t~ sin θ \quad\quad\quad (a)
in which ɸ is the angle to the element of area and t is the thickness of the section. Thus, the shear stress \tau at section bb is
\quad\quad\quad τ = \frac{V_yQ_z}{I_zt} = \frac{V_yr^2 sin θ}{I_z}\quad\quad\quad (b)
Substituting I_{z}\,=\,\pi r^{3}t/2 (see Case 22 or Case 23 of Appendix E) gives
\quad\quad \tau={\frac{2V_{y}\ \sin\theta}{\pi r t}}\quad\quad\quad\quad (6-85)
When θ = 0 ~ or ~ θ = \pi , this expression gives \tau = 0, as expected. When θ = \pi/2, it gives the maximum shear stress.
3. Analyze:
Location of shear center: The resultant of the shear stresses must be the vertical shear force V_y. Therefore, the moment M_0 of the shear stresses about the center O must equal the moment of the force V_y about that same point:
\quad\quad\quad M_0 = V_y e \quad\quad\quad (c)
To evaluate M_0, begin by noting that the shear stress τ acting on the element of area dA (Fig. 6-45b) is
\quad\quad\quad τ = \frac{2V_y sin ɸ}{\pi rt}
as found from Eq. (6-85). The corresponding force is \tau ~dA, and the moment of this force is
\quad\quad\quad dM_0 = r(τ dA) = \frac{2V_y sin ɸ dA}{\pi t}
Since dA = tr  dɸ, this expression becomes
\quad\quad\quad dM_0 = \frac{2r V_y sin ɸ dɸ}{\pi}
Therefore, the moment produced by the shear stresses is
\quad M_0 = \int dM_0 = \int\limits_{0}^{\pi}\frac{2r V_y sin ɸ dɸ}{\pi} = \frac{4rV_y}{\pi} \quad\quad (d)
It follows from Eq. (c) that the distance e to the shear center is
\quad\quad\quad e = \frac{M_0}{V_y} = \frac{4r}{\pi} ≈ 1.27 r \quad\quad (6-86)
4. Finalize: This result shows that the shear center S is located outside of the semicircular section.
Note: The distance from the center O of the semicircle to the centroid C of the cross section (Fig. 6-45a) is 2r/\pi  (from Case 23 of Appendix E), which is one-half of the distance e. Thus, the centroid is located midway between the shear center and the center of the semicircle.
The location of the shear center in a more general case of a thin-walled circular section is determined in Prob. 6.9-13.