Handbook of Civil Engineering Calculations

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. 426, a free-body diagram of each member.


3. Evaluate the deflection of the beams
   Evaluate, in terms of P, the deflections yl2 and ^ 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 mo-
   ment. Then compute the corresponding flexural stress.

THEOREMOFTHREEMOMENTS

For the two-span beam in Fig. 43a, compute the reactions at the supports. Apply the theo-
rem of three moments to arrive at the results.

Calculation Procedure:

1. Using the bending-moment equation, determine MB
   Figure 436 represents a general case. For a prismatic beam, the bending moments at the
   three successive supports are related by M 1 L 1 + 2M 2 (L 1 + L 2 ) + M 3 L 2 -^1 ^w 1 L? -^1 Aw 2 Lq -
   P 1 L^k 1 - k) - P 2 LKk 2 - K 2 ). Substituting in this equation gives M 1 = M 3 = O; 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 N); P 2 = 10 kips (44.5 N); Jt 1 = 0.5; k 2 = 0.4; 2M 5 (IO + 15) = -Vi(2)(10)^3
   -^1 X 4 (S)(IS)^3 - 6(10)^2 (0.5 - 0.125) - 10(15)^2 (0.4 - 0.064); MB = -80.2 ft-kips (-108.8
   kN-m).


2. Draw a free-body diagram of each span
   Figure 43 c shows the free-body diagrams.