Handbook of Civil Engineering Calculations

Calculation Procedure:

1. Draw the free-body diagram of the beam
   Draw the diagram in Fig. 4lb. Consider this as a simply supported member carrying a 50-
   kip (222.4-kN) load at D and an upward load R 8 at its center.


2. Evaluate the deflection
   Evaluate the deflection at B by applying the equations presented for cases 7 and 8 in the
   AISC Manual With respect to the 50-kip (222.4-kN) load, b = 1 ft (2.1 m) and x = 14 ft
   (4.3 m). If y is in inches and R 8 is in pounds, y = 50,000(7)(14)(28^2 - 72 - 142 )1728/
   [6(35)(10)^9 28] -^(28)^3 1728/[48(35)(10)^9 ] = 0.776 - (2.26/10^5 )^.


3. Express the deflection in terms of the spring constant
   The deflection at B is, by proportion, yl\ = 1^/100,000; y = #g/100,000.


4. Equate the two deflection expressions, and solve for the
   upward load
   Thus RB/W^5 = 0.776 - (2.26/10^5 )^; R 8 = 0.776(10)^5 /3.26 = 23,800 Ib (105,862.4 N).


5. Calculate the reactions RA and R 0 by taking moments
   We have 2MC = 2%RA - 50,000(21) + 23,800(14) = O; RA = 25,600 Ib (113,868.8 N);
   SM 4 = 50,000(7) -23,800(14) - 287?c = O; Rc = 600 Ib (2668.8 N).


6. Construct the shear and moment diagrams
   Construct these diagrams as shown in Fig. 41. Then Mn = 7(25,600) = 179,200 lb-ft
   (242,960 NUI); M 8 =179,200 - 7(24,400) = 8400 lb-ft (11,390.4 N-m).

MAXIMUM BENDING STRESS IN BEAMS

JOINTLY SUPPORTING A LOAD

In Fig. 420, a W16 x 40 beam and a W12 x 31 beam cross each other at the vertical line
V, the bottom of the 16-in (406.4-mm) beam being % in (9.53 mm) above the top of the
12-in (304.8-mm) beam before the load is applied. Both members are simply supported.

FIGURE 42. Load carried by two beams.