Page 81 - Handbook of Civil Engineering Calculations, Second Edition

1.64 STRUCTURAL STEEL ENGINEERING AND DESIGN

b. Remove the fixed support at D and add the span DE of zero length, with a hinged sup-
port at E.
For the interval BD, the transformed beam is then identical in every respect with the
actual beam.
2. Apply the equation for the theorem of three moments
Consider span BC as span 1 and CD as span 2. For the 5-kip (22.2-kN) load, k 2 12/16
0.75; for the 10-kip (44.5-kN) load, k 2 8/16 0.5. Then 12(10) 2M C (10 16)
2
2
3
16M D /4(4)(10) 5(16) (0.75 0.422) 10(16) (0.5 0.125). Simplifying
1
gives 13M C 4M D 565.0, Eq. a.
3. Apply the moment equation again
Considering CD as span 1 and DE as span 2, apply the moment equation again. Or, for the
5-kip (22.2-kN) load, k 1 0.25; for the 10-kip (44.5-kN) load, k 1 0.5. Then 16M C
2 2
2M D (16 0) 5(16) (0.25 0.016) 10(16) (0.50 0.125). Simplifying yields M C
2M D 78.7, Eq. b.
4. Solve the moment equations
Solving Eqs. a and b gives M C 37.1 ft·kips ( 50.30 kN·m); M D 20.8 ft·kips
( 28.20 kN·m).
5. Determine the reactions by using a free-body diagram
Find the reactions by drawing a free-body diagram of each span and taking moments with
respect to each support. Thus R B 20.5 kips (91.18 kN); R C 32.3 kips (143.67 kN);
R D 5.2 kips (23.12 kN).

BENDING-MOMENT DETERMINATION BY
MOMENT DISTRIBUTION

Using moment distribution, determine the bending moments at the supports of the member in
Fig. 45. The beams are rigidly joined at the supports and are composed of the same material.

Calculation Procedure:

1. Calculate the flexural stiffness of each span
Using K to denote the flexural stiffness, we see that K I/L if the far end remains fixed
during moment distribution; K 0.75I/L if the far end remains hinged during moment
distribution. Then K AB 270/18 15; K BC 192/12 16; K CD 0.75(240/20) 9.
Record all the values on the drawing as they are obtained.
2. For each span, calculate the required fixed-end moments at
those supports that will be considered fixed
These are the external moments with respect to the span; a clockwise moment is consid-
ered positive. (For additional data, refer to cases 14 and 15 in the AISC Manual.) Then
2
2
M AB wL /12 2(18) /12 54.0 ft·kips ( 73.2 kN·m); M BA 54.0 ft·kips
(73.22 kN·m). Similarly, M BC 48.0 ft·kips ( 65.1 kN·m); M CB 48.0 ft·kips (65.1
2
kN·m); M CD 24(15)(5)(15 20)/[2(20) ] 78.8 ft·kips ( 106.85 kN·m).
3. Calculate the unbalanced moments
Computing the unbalanced moments at B and C yields the following: At B, 54.0 48.0
6.0 ft·kips (8.14 kN·m); at C, 48.0 78.8 30.8 ft·kips ( 41.76 kN·m).