(a) Transverse section (b) Force diagram
FIGURE 26
Calculation Procedure:
- Compute the left reaction and the section at which the
shear is zero
The left reaction R 1 =^1 X 2 (IO)(SOO) + 1600(2.5/10) = 2900 Ib (12,899.2 N). The section A
at which the shear is zero is x = 2900/500 = 5.8 ft (1.77 m).
- Compute the maximum moment
Use the relation M 4 = ^(290O)(S-S) = 8410 lb-ft (11,395.6 N-m) = 100,900 lb-in
(11,399.682 N-m).
- Locate the centroidal axis of the section
Use the AISC Manual for properties of the trapezoid. Oryt = (9/3)[(2 * 6 + 3)1(6 + 3)] =
5 in (127 mm); yb = 4m (101.6 mm).
- Compute the moment of inertia of the section
Using the AISC Manual, I= (9^3 /36)[(6^2 + 4x6x3 + 32 )/(6 + 3)] = 263.3 in^4 (10,959.36
cm^4 ).
- Compute the stresses in the beam
Use the relation/= MyII, where/= bending stress in a given fiber, lb/in^2 (kPa); y = dis-
tance from neutral axis to given fiber, in. Thus/op = 100,900(5)/263.3 = 1916-lb/in^2
(13,210.8-kPa) compression,/bottom = 100,900(4)/263.3 = 1533-lb/in^2 (10,570.0-kPa) ten-
sion.
In general, the maximum bending stress at a section where the moment is M is given
by/= Mc/7, where c = distance from the neutral axis to the outermost fiber, in (mm). For
a section that is symmetric about its centroidal axis, it is convenient to use the section
modulus S of the section, this being defined as S = Uc. Then/= MIS.
ANALYSIS OF A BEAM ON MOVABLE
SUPPORTS
The beam in Fig. 270 rests on two movable supports. It carries a uniform live load of w
Ib/lin ft and a uniform dead load of 0.2w Ib/lin ft. If the allowable bending stresses in ten-
sion and compression are identical, determine the optimal location of the supports.
