Page 79 - Handbook of Civil Engineering Calculations, Second Edition

1.62 STRUCTURAL STEEL ENGINEERING AND DESIGN

A column bearing on the 16-in. (406.4-mm) beam transmits a load of 15 kips (66.72 kN)
at the indicated location. Compute the maximum bending stress in the 12-in. (304.8-mm)
beam.

Calculation Procedure:

1. Determine whether the upper beam engages the lower beam
To ascertain whether the upper beam engages the lower one as it deflects under the 15-kip
(66.72-kN) load, compute the deflection of the 16-in. (406.4-mm) beam at V if the 12-in.
(304.8-mm) beam were absent. This distance is 0.74 in. (18.80 mm). Consequently, the
gap between the members is closed, and the two beams share the load.
2. Draw a free-body diagram of each member
Let P denote the load transmitted to the 12-in. (304.8-mm) beam by the 16-in. (406.4-mm)
beam [or the reaction of the 12-in. (304.8-mm) beam on the 16-in. (406.4-mm) beam].
Draw, in Fig. 42b, a free-body diagram of each member.
3. Evaluate the deflection of the beams
Evaluate, in terms of P, the deflections y 12 and y 16 of the 12-in. (304.8-mm) and 16-in.
(406.4-mm) beams, respectively, at line V.
4. Express the relationship between the two deflections
Thus, y 12 y 16 0.375.
5. Replace the deflections in step 4 with their values as obtained
in step 3
After substituting these deflections, solve for P.
6. Compute the reactions of the lower beam
Once the reactions of the lower beam are computed, obtain the maximum bending
moment. Then compute the corresponding flexural stress.

THEOREM OF THREE MOMENTS

For the two-span beam in Fig. 43 and 44, compute the reactions at the supports. Apply the
theorem of three moments to arrive at the results.

Calculation Procedure:

1. Using the bending-moment equation, determine M B
Figure 43 represents a general case. For a prismatic beam, the bending moments at the three
3
3
successive supports are related by M 1 L 1 2M 2 (L 1 L 2 ) M 3 L 2 /4w 1 L 1 /4w 2 L 2
1
1
3
2
3
2
P 1 L 1 (k 1 k 1 ) P 2 L 2 (k 2 k 2 ). Substituting in this equation gives M 1 M 3 0; L 1 10 ft
(3.0 m); L 2 15 ft (4.6 m); w 1 2 kips/lin ft (29.2 kN/m); w 2 3 kips/lin ft (43.8 kN/m);
P 1 6 kips (26.7 kN); P 2 10 kips (44.5 kN); k 1 0.5; k 2 0.4; 2M B (10 15)
2
2
3
3
1 /4(2)(10) /4(3)(15) 6(10) (0.5 0.125) 10(15) (0.4 0.064); M B 80.2
1
ft·kips ( 108.8 kN·m).
2. Draw a free-body diagram of each span
Figure 43 shows the free-body diagrams.