Question 6.7: A thin-walled semicircular cross section of radius r and thi...
A thin-walled semicircular cross section of radius r and thickness t is shown in Fig. 6-36a. Determine the distance e from the center O of the semicircle to the shear center S.
We know immediately that the shear center is located somewhere on the axis of symmetry (the z axis). To determine the exact position, we 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. 6-36b).
Shear stresses. The first step is to determine the shear stresses τ acting on the cross section (Fig. 6-36b). We 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, we identify an element of area dA (shown shaded in the figure) and integrate as follows:
Q_{z}=\int{y dA}=\int_{0}^{\theta}{(r\cos\phi)(tr d\phi)}=r^{2}t\sin\theta (e)
in which Φ is the angle to the element of area and t is the thickness of the section. Thus, the shear stress τ at section bb is
\tau=\frac{V_{y}Q_{z}}{I_{z}t}=\frac{V_{y}r^{2}\sin\theta}{I_{z}} (f)
Substituting I_{z} = πr³t/2 (see Case 22 or Case 23 of Appendix D), we get
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Thin circular ring (Origin of axes at center) Approximate formulas for case when t is small A=2\pi rt=\pi dt I_{x}=I_{y}=\pi r^{3}t=\frac{\pi d^{3}t}{8}
I_{xy}=0 I_{P}=2\pi r^{3}t=\frac{\pi d^{3}t}{4} |
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Thin circular arc (Origin of axes at center of circle) Approximate formulas for case when t is small β = angle in radians (Note: For a semicircular arc, β = π/2.)
A = 2βrt \overline{y}=\frac{r\sin\beta}{\beta}
I_{x} = r³t(β + sin β cos β) I_{y} = r³t(β – sin β cos β)
I_{xy} = 0 I_{BB}=r^{3}t\left(\frac{2\beta + \sin2\beta}{2}-\frac{1 – \cos2\beta}{\beta}\right) |
\tau=\frac{2V_{y}\sin\theta}{\pi rt} (6-72)
When θ = 0 or θ = π, this expression gives τ = 0, as expected. When θ = π/2, it gives the maximum shear stress.
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:
M_{0}=V_{y}e (g)
To evaluate M_{0}, we begin by noting that the shear stress τ acting on the element of area dA (Fig. 6-36b) is
\tau=\frac{2V_{y}\sin\phi}{\pi rt}as found from Eq. (6-72). The corresponding force is τ dA, and the moment of this force is
dM_{0}=r(\tau dA)=\frac{2V_{y}\sin\phi dA}{\pi t}Since dA = trdΦ, this expression becomes
dM_{0}=\frac{2rV_{y}\sin\phi d\phi}{\pi}Therefore, the moment produced by the shear stresses is
M_{0}=\int{dM_{0}}=\int_{0}^{\pi}\frac{2rV_{y}\sin\phi d\phi}{\pi}=\frac{4rV_{y}}{\pi} (h)
It follows from Eq. (g) that the distance e to the shear center is
e=\frac{M_{0}}{V_{y}}=\frac{4r}{\pi}\approx 1.27r (6-73)
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. 6-36a) is 2r/π (from Case 23 of Appendix D), 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 Problem 6.9-12.
