Page 66 - Handbook of Civil Engineering Calculations, Second Edition

STATICS, STRESS AND STRAIN, AND FLEXURAL ANALYSIS 1.49

shearing force at any section must pass through a particular point on the centroidal axis
designated as the shear, or flexural, center.
Cut the beam at section 2, and consider the left portion of the beam as a free body. In
Fig. 31b, indicate the resisting shearing forces V 1 , V 2 , and V 3 that the right-hand portion of
the beam exerts on the left-hand portion at section 2. Obtain the directions of V 1 and V 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. 31c. 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 M 0 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.
2. Compute the shear flow
Determine the shear flow at E and F (Fig. 31) by setting Q in q VQ/I equal to the static
moment of the overhanging portion of the flange. (For convenience, use the dimensions
3
2
1
to the centerline of the web and flange.) Thus I /12(0.10)(16) 2(8)(0.10)(8) 137
3
3
in (5702.3 cm ); Q BE 5(0.10)(8) 4.0 in (65.56 cm ); Q FG 3(0.10)(8) 2.4 in 3
4
4
(39.34 cm ); q E VQ BE /I 10,000(4.0)/137 292 lb/lin in (51,137.0 N/m); q F
3
10,000(2.4)/137 175 lb/lin in (30,647.2 N/m).
3. Compute the shearing forces on the transverse section
Since the shearing stress varies linearly across the flange, V 1 /2(292)(5) 730 lb
1
1
(3247.0 N); V 2 /2(175)(3) 263 lb (1169.8 N); V 3 P 10,000 lb (44,480 N).
4. 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(V 2 V 1 ) 0, or 10,000e
16(263 730) 0; e 0.747 in. (18.9738 mm).
5. Verify the computed values
Check the computed values of q E and q F by considering the bending stresses directly. Ap-
ply the equation f Vy/I, where f increase in bending stress per unit distance along
the span at distance y from the neutral axis. Then f 10,000(8)/137 584 lb/(sq.in.·in)
(158.52 MPa/m).
In Fig. 31c, set AB 1 in. (25.4 mm). Then q E 584(5)(0.10) 292 lb/lin in
(51,137.0 N/m); q F 584(3)(0.10) 175 lb/lin in (30,647.1 N/m).
Although a particular type of beam (cantilever) was selected here for illustrative
purposes and a numeric value was assigned to the vertical shear, note that the value of
e is independent 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 OF A CIRCULAR FLAT PLATE

A circular steel plate 2 ft (0.61 m) in diameter and /2 in. (12.7 mm) thick, simply support-
1
ed along its periphery, carries a uniform load of 20 lb/sq.in. (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.