Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

SHEAR CENTERS OF
THIN-wALLEd OPEN
SECTIONS
377
P
P
(a) External load P applied horizontally
through the centroid
(b) Cantilever beam bends without twisting
FIGURE 9.29 External load acting in a plane of symmetry.
are two equal resultant shear forces that act in opposite directions above and below the
axis of symmetry.
In conclusion, as long as the external loads act through the shear center, the beam will
bend without twisting. When this requirement is met, the stresses in the beam can be determined
from the flexure formula.
Determination of the Shear center Location
The location of the shear center for an unsymmetrical shape is computed as follows:
1. Determine how the shear “flows” in the various portions of the cross section.
2. Determine the distribution of shear flow q for each portion of the cross section from
the shear flow equation q = VQ/I. Convert the shear flow into a force resultant by
integrating q along the length of the cross-sectional element. The shear flow q will vary
(a) linearly in elements that are perpendicular to the direction of the internal shear
force V and (b) parabolically in elements that are parallel to or inclined toward the
direction of V.
3. Alternatively, determine the distribution of shear stress τ from the shear stress equation
τ = VQ/It and convert the shear stress into a force resultant by integrating τ over the
area of the cross-sectional element.
4. Sketch the shear force resultants that act in each element of the cross section.
5. Determine the shear center location by summing moments about an arbitrary point (e.g.,
point B) on the cross section. Choose a convenient location for point B—one that
eliminates as many force resultants from the moment equilibrium equation as possible.
6. Study the direction of rotation of the shear forces, and place the external force P at an
eccentricity e from point B so that the direction of the moment Pe is opposite to that
caused by the resultant shear forces.
7. Sum moments about point B, and solve for the eccentricity e.
8. If the cross section has an axis of symmetry, then the shear center lies at the point
where this axis intersects the line of action of the external load. If the shape has no
axes of symmetry, then rotate the cross section 90° and repeat the process to obtain
another line of action for the external loads. The shear center lies at the intersection of
these two lines.