- 5.1 Introduction
- 5.2 Pure Bending of Beams of Symmetrical Cross Section
- 5.3 Pure Bending of Beams of Asymmetrical Cross Section
- 5.4 Bending of A Cantilever of Narrow Section
- 5.5 Bending of a Simply Supported Narrow Beam
- 5.6 Elementary Theory of Bending
- 5.7 Normal and Shear Stresses
- 5.8 Effect of Transverse Normal Stress
- 5.9 Composite Beams
- 5.10 Shear Center
- 5.11 Statically Indeterminate Systems
- 5.12 Energy Method for Deflections
- 5.13 Elasticity Theory
- 5.14 Curved Beam Formula
- 5.15 Comparison of the Results of Various Theories
- 5.16 Combined Tangential and Normal Stresses
- References
- Problems
This chapter is from the book
This chapter is from the book
This chapter is from the book
5.10 Shear Center
Given any cross-sectional configuration, one point may be found in the plane of the cross section through which the resultant of the transverse shearing stresses passes. A transverse load applied on the beam must act through this point, called the shear center or flexural center, if no twisting is to occur [Ref. 5.3]. The center of shear is sometimes defined as the point in the end section of a cantilever beam at which an applied load results in bending only. When the load does not act through the shear center, in addition to bending, a twisting action results (Section 6.1).
The location of the shear center is independent of the direction and magnitude of the transverse forces. For singly symmetrical sections, the shear center lies on the axis of symmetry, while for a beam with two axes of symmetry, the shear center coincides with their point of intersection (also the centroid). It is not necessary, in general, for the shear center to lie on a principal axis, and it may be located outside the cross section of the beam.
5.10.1 Thin-Walled Open Cross Sections
For thin-walled sections, the shearing stresses are taken to be distributed uniformly over the thickness of the wall and directed so as to parallel the boundary of the cross section. If the shear center S for the typical section of Fig. 5.18a is required, we begin by calculating the shear stresses by means of Eq. (5.39). The moment Mx of these stresses about arbitrary point A is then obtained. Since the external moment attributable to Vy about A is Vye, the distance between A and the shear center is given by
)
)
Figure 5.18. Shear centers.
If the force is parallel to the z axis rather than the y axis, the position of the line of action may be established in the manner discussed previously. If both Vy and Vz exist, the shear center is located at the intersection of the two lines of action.
The determination of Mx is simplified by propitious selection of point A, such as in Fig. 5.18b. There, the moment Mx of the shear forces about A is zero; point A is also the shear center. For all sections consisting of two intersecting rectangular elements, the same situation exists.
For thin-walled box beams (with boxlike cross section), the point or points in the wall where the shear flow q = 0 (or τxy = 0) is unknown. Here, shear flow is represented by the superposition of transverse and torsional flow (see Section 6.8). Hence, the unit angle of twist equation, Eq. (6.23), along with q = VQ/I, is required to find the shear flow for a cross section of a box beam. The analysis procedure is as follows: First, introduce a free edge by cutting the section open; second, close it again by obtaining the shear flow that makes the angle of twist in the beam zero [Refs. 5.4 through 5.6].
In the following examples, the shear center of an open, thin-walled section is determined for two typical situations. In the first, the section has only one axis of symmetry; in the second, there is an asymmetrical section.
EXAMPLE 5.9 SHEARING STRESS DISTRIBUTION IN A CHANNEL SECTION
Locate the shear center of the channel section loaded as a cantilever (Fig. 5.19a). Assume that the flange thicknesses are small when compared with the depth and width of the section.
)
Figure 5.19. Example 5.9. (a) Cantilever beam with a concentrated load at the free end; (b) an element of the upper flange; (c) shear distribution; (d) location of the shear center S.
Solution The shearing stress in the upper flange at any section nn will be found first. This section is located a distance s from the free edge m, as shown in the figure. At m, the shearing stress is zero. The first moment of area st1 about the z axis is Qz = st1h. The shear stress at nn, from Eq. (5.39), is thus
)
The direction of τ along the flange can be determined from the equilibrium of the forces acting on an element of length dx and width s (Fig. 5.19b). Here the normal force N = t1sσx, owing to the bending of the beam, increases with dx by dN. Hence, the x equilibrium of the element requires that τt1·dx must be directed as shown in the figure. This flange force is directed to the left, because the shear forces must intersect at the corner of the element.
The distribution of the shear stress τxy on the flange, as Eq. (a) indicates, is linear with s. Its maximum value occurs at
)
Similarly, the value of stress τxy at the top of the web is
)
The stress varies parabolically over the web, and its maximum value is found at the neutral axis. Figure 5.19c sketches the shear stress distribution in the channel. As the shear stress is linearly distributed across the flange length, from Eq. (b), the flange force is expressed by
)
Symmetry of the section dictates that F1 = F3 (Fig. 5.19d). We will assume that the web force F2 = p, since the vertical shearing force transmitted by the flange is negligibly small, as shown in Example 5.3. The shearing force components acting in the section must be statically equivalent to the resultant shear load P. Thus, the principle of the moments for the system of forces in Fig. 5.19d or Eq. (5.56), applied at A, yields Mx = pe = 2F1h. Substituting F1 from Eq. (d) into this expression, we obtain
Since for the usual channel section t1 is small in comparison to b or h, the simplified moment of inertia has the following form:
The shear center is, in turn, located by the expression
)
Comments Note that e depends on only the section dimensions. Examination reveals that e may vary from a minimum of zero to a maximum of b/2. A zero or near-zero value of e corresponds to either a flangeless beam (b = 0, e = 0) or an especially deep beam (h≫b. The extreme case, e = b/2, is obtained for an infinitely wide beam.
EXAMPLE 5.10 SHEAR FLOW IN AN ASYMMETRICAL CHANNEL SECTION
Locate the shear center S for the asymmetrical channel section shown in Fig. 5.20a. All dimensions are in millimeters. Assume that the beam thickness t = 125 mm is constant.
)
Figure 5.20. Example 5.10. (a) Portion of a beam with a channel cross section; (b) shear flow; (c) location of the shear center S.
Solution The centroid C of the section is located by 
and 
with respect to nonprincipal axes z and y. By performing the procedure given in Example 5.1, we obtain 
, Iy = 4765.62 mm4, Iz = 21,054.69 mm4, and Iyz = 3984.37 mm4. Equation (5.18) then yields the direction of the principal axis x′, y′ as θp = 13.05°, and Eq. (5.19) gives the principal moments of inertia as Iy′ = 3828.12 mm4, Iz′ = 21,953.12 mm4 (Fig. 5.20a).
Let us now assume that a shear load Vy′ is applied in the y′, z′ plane (Fig. 5.20b). This force may be considered the resultant of force components F1, F2 and F3 acting in the flanges and web in the directions indicated in the figure. The algebra will be minimized if we choose point A, where F2 and F3 intersect, in finding the line of action of Vy′ by applying the principle of moments. To do so, we need to determine the value of F1 acting in the upper flange. The shear stress τxz in this flange, from Eq. (5.39), is
)
where s is measured from right to left along the flange. Note that Qz′ the bracketed expression, is the first moment of the shaded flange element area with respect to the Z′ axis. The constant 19.55 is obtained from the geometry of the section. After substituting the numerical values and integrating Eq. (f), the total shear force in the upper flange is found to be
)
Application of the principle of moments at A gives Vy′ez′ = 37.5F1. Introducing F1 from Eq. (g) into this equation, the distance ez′, which locates the line of action of Vy′ from A, is
)
Next, assume that the shear loading Vz′ acts on the beam (Fig. 5.20c). The distance ey′ may be obtained as in the situation just described. Because of Vz′ the force components F1 to F4 will be produced in the section. The shear stress in the upper flange is given by
)
The quantity Qy′ represents the first moment of the flange segment area with respect to the y′ axis, and 12.05 is found from the geometry of the section. The total force F1 in the flange is
)
The principle of moments applied at A,Vz′ey′ = 37.5F1 = 7.65Vz′ leads to
)
Thus, the intersection of the lines of action of Vy′ and Vz′, and ez′ and ey′ locates the shear center S of the asymmetrical channel section.
5.10.2 Arbitrary Solid Cross Sections
The preceding considerations can be extended to beams of arbitrary solid cross section, in which the shearing stress varies with both cross-sectional coordinates y and z. For these sections, the exact theory can, in some cases, be successfully applied to locate the shear center. Examine the section of Fig. 5.18c subjected to the shear force Vz, which produces the stresses indicated. Denote y and z as the principal directions. The moment about the x axis is then
)
Vz must be located a distance e from the z axis, where e = Mx/vz.