centroidal axis. If bending of the beam is not to be accompanied by torsion, the vertical
shearing force at any section must pass through a particular point on the centroidal axis
designated as the shear, orflexural, center.
Cut the ,beam at section 2, and consider the left portion of the beam as a free body. In
Fig. 3 Ib, indicate the resisting shearing forces K 1 , K 2 , and K 3 that the right-hand portion of
the beam exerts on the left-hand portion at section 2. Obtain the directions of K 1 and K 2
this way: Isolate the segment of the beam contained between sections 1 and 2; then isolate
a segment ABDC of the top flange, as shown in Fig. 3 Ic. Since the bending stresses at
section 2 exceed those at section 1, the resultant tensile force T 2 exceeds T 1. The resisting
force on CD is therefore directed to the left. From the equation of equilibrium SM = O it
follows that the resisting shears on AC and BD have the indicated direction to constitute a
clockwise couple.
This analysis also reveals that the shearing stress varies linearly from zero at the edge
of the flange to a maximum value at the juncture with the web.
- Compute the shear flow
Determine the shear flow at E and F (Fig. 31) by setting Q'mq= VQII equal to the static
moment of the overhanging portion of the flange. (For convenience, use the dimensions
to the centerline of the web and flange.) Thus /=
1
Xw(O.! O)(16)
3
- 2(8)(0.10)(S)
2
= 137 in
4
(5702.3 cm^4 ); QBE = 5(0.10)(8) = 4.0 in^3 (65.56 cm^3 ); QFG•= 3(0.10)(8) = 2.4 in^3 (39.34
cm^3 ); ^= VQ 8 JI= 10,000(4.0)/!37 = 292 Ib/lin in (5 U37.0 N/m); ?F = 10,000(2.4)/!37
= 175 Ib/lin in (30,647.2 N/m).
- Compute the shearing forces on the transverse section
Since the shearing stress varies linearly across the flange, K 1 = H(292)(5) = 730 Ib
(3247.0 N); K 2 =^1 A(IlS)Q) = 263 Ib (1169.8 N); K 3 = P = 10,000 Ib (44,480 N). - Locate the shear center
Take moments of all forces acting on the left-hand portion of the beam with respect to a
longitudinal axis through the shear center O. Thus V 3 e + 16(K 2 - K 1 ) = O, or 10,00Oe +
16(263 - 730) - O; e = 0.747 in (18.9738 mm). - Verify the computed values
Check the computed values ofqE and qF by considering the bending stresses directly. Ap-
ply the equation A/= VyII 9 where A/= increase in bending stress per unit distance along
the span at distance y from the neutral axis. Then A/= 10,000(8)/137 = 584 Ib/(in^2 -in)
(158.52 MPa/m).
In Fig. 31c, set AB = 1 in (25.4 mm). Then qE = 584(5)(0.10) = 292 Ib/lin in (51,137.0
N/m); qF = 584(3)(0.10) = 175 Ib/lin in (30,647.1 N/m).
Although a particular type of beam (cantilever) was selected here for illustrative pur-
poses and a numeric value was assigned to the vertical shear, note that the value of e is in-
dependent of the type of beam, form of loading, or magnitude of the vertical shear. The
location of the shear center is a geometric characteristic of the transverse section.
BENDING OFA CIRCULAR FLATPLATE
A circular steel plate 2 ft (0.61 m) in diameter and^1 A in (12.7 mm) thick, simply support-
ed along its periphery, carries a uniform load of 20 lb/in^2 (137.9 kPa) distributed over the
entire area. Determine the maximum bending stress and deflection of this plate, using
0.25 for Poisson's ratio.