Thin walled l section shear centre is located at which point
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Having found the solution to symmetrical and asymmetrical bending, in this section we find where the load has to be applied so that it produces no torsion.
Shear center is defined as the point about which the external load has to be applied so that it produces no twisting moment.
Recall from equation (8.7) the torsional moment due to the shear force σxy and σxz about the origin is,
(8.210)
Since, ∫ aσxzdydz = V z and ∫ aσxydydz = V y, the moment about some other point (ysc,zsc) would be,
(8.211)
If this point (ysc,zsc) is the shear center, then Mxsc = 0. Thus, we have to find y sc and zsc such that,
(8.212)
holds. We have two unknowns but only one equation. Hence, we cannot find ysc and zscuniquely, in general. If the loading is such that only shear force V y is present, then
(8.213)
Similarly, if V y = 0,
(8.214)
Equations (8.213) and (8.214) are used to find the coordinates of the shear center with respect to the chosen origin of the coordinate system, which for homogeneous sections is usually taken as the centroid of the cross section. Thus, the point that (ysc,zsc) are the coordinates of the shear center from the origin of the chosen coordinate system which in many cases would be the centroid of the section cannot be overemphasized. In the case of thin walled sections which develop shear stresses tangential to the cross section, σxy = -τsin(θ) and σxz = τ cos(θ), where τ is the magnitude of the shear stress and θ is the angle the tangent to the cross section makes with the z direction.
By virtue of the shear stress depending linearly on the shear force (see equations (8.43) and (8.207)), it can be seen that the coordinates of the shear center is a geometric property of the section.
Shear center is defined as the point about which the external load has to be applied so that it produces no twisting moment.
Recall from equation (8.7) the torsional moment due to the shear force σxy and σxz about the origin is,
(8.210)
Since, ∫ aσxzdydz = V z and ∫ aσxydydz = V y, the moment about some other point (ysc,zsc) would be,
(8.211)
If this point (ysc,zsc) is the shear center, then Mxsc = 0. Thus, we have to find y sc and zsc such that,
(8.212)
holds. We have two unknowns but only one equation. Hence, we cannot find ysc and zscuniquely, in general. If the loading is such that only shear force V y is present, then
(8.213)
Similarly, if V y = 0,
(8.214)
Equations (8.213) and (8.214) are used to find the coordinates of the shear center with respect to the chosen origin of the coordinate system, which for homogeneous sections is usually taken as the centroid of the cross section. Thus, the point that (ysc,zsc) are the coordinates of the shear center from the origin of the chosen coordinate system which in many cases would be the centroid of the section cannot be overemphasized. In the case of thin walled sections which develop shear stresses tangential to the cross section, σxy = -τsin(θ) and σxz = τ cos(θ), where τ is the magnitude of the shear stress and θ is the angle the tangent to the cross section makes with the z direction.
By virtue of the shear stress depending linearly on the shear force (see equations (8.43) and (8.207)), it can be seen that the coordinates of the shear center is a geometric property of the section.
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