### STRUCTURES INDETERMINATE BEAMS

Indeterminate beams include beams with fixed-end (moment resistant) supports and beams of more than two supports, referred to as continuous beams. The design of  statically indeterminate beams cannot be done by static equations alone. However, bending coefficients, derived by mechanics, may be used for analysis of typical beams.

The bending moment is computed, multiplying the bending coefficients by the total load W and span L between supports.  For continuous beams, the method is limited to beams  of equal spans for all bays.  The coefficients here assume all bays are loaded.

Coefficients for alternating live load on some bays and combined dead load plus live load  on others, which may result in greater bending moments, are in Appendix A.  Appendix A  also has coefficients for other load conditions, such as various point loads. The equation  for bending moments by bending coefficients is:

M = C L W

M = bending moment
C = bending coefficient
L = span between supports
W = w L (total load per bay)
w = uniform load in plf (pounds / linear foot

1 Simple beam
2  Fixed-end beam (combined positive plus negative moments equal the simple-beam moment)
3 Two-span beam
4 Three-span beam

### STRUCTURES EXAMPLE BEAM ANALYSIS

Reactions

Shear

Find x, where shear = 0 and bending = maximum:

Vbr-w2 x = 0;  x = Vbr/w2 = 2000/200      x = 10 ft

Moment

Section modulus

Bending stress

Shear stress
Note: stress is figured, using absolute maximum bending and shear, regardless if positive  or negative.  Lumber sizes are nominal, yet  actual sizes are used for computation.   Actual sizes are ½ in. less for lumber up to 6 in. nominal and ¾ in. less for larger sizes:  4x8 nominal is 3½x7¼ in. actual.

Note: in the above two beams shear stress is more critical (closer to the allowable stress)  than bending stress, because negative cantilever moments partly reduces positive  moments.

### STRUCTURES EXAMPLE BEAM DESIGN

V   Shear diagram.
M  Bending diagram.
∆   Deflection diagram.
I    Inflection point (change from + to - bending).

Reactions

R= 400plf (24)/2      R = 4800 lbs

Shear

Moment

Try 4x10 beam

Bending stress

f b = Mmax/S= 5000 (12)/50      f b = 1200 psi
1200 < 1450, ok

Shear stress

f v = 1.5V/bd=1.5(2800)/[3.5(9.25)]       f v = 130 psi
130 > 95, not ok

Try 6x10 beam
f v = 1.5V/bd=1.5 (2800)/[5.5 (9.25)]      f v= 83 psi
83 < 95, ok

Note: increased beam width is most effective to reduce shear stress; but increased depth  is most effective to reduce bending stress.

### STRUCTURES - AREA METHOD FOR BEAM DESIGN

The area method for beam design simplifies computation of shear forces and bending moments and is derived, referring to the following diagrams:
1  Load diagram on beam
2 Beam diagram
3 Shear diagram
4 Bending diagram

The area method may be stated:

•  The shear at any point n is equal to the shear at point m plus  the area of the load diagram between m and n.

•  The bending moment at any point n is equal to the moment at point m plus the area of the shear diagram between m and n.

The shear force is derived using vertical equilibrium:

∑ V = 0;   Vm - w x - Vn = 0;  solving for Vn

Vn = Vm-wx

where w x is the load area between m and n (downward load w = negative).

The bending moment is derived using moment equilibrium:

∑ M = 0;    Mm + Vmx – w x x/2 - Mn = 0;  solving for Mn

where Vmx – wx^2/2 is the shear area between m and n, namely, the rectangle    Vm x less the triangle w x^2/2.  This relationship may also be stated as Mn = Mm + Vx, where V is the average shear between m and n.

By the area method moments are usually equal to the area of one or more rectangles and/or triangles.  It is best to first compute and draw the shear diagram and then compute the moments as the area of the shear diagram.

From the diagrams and derivation we may conclude:

•  Positive shear implies increasing bending moment.
•  Zero shear (change from + to -) implies peak bending moment (useful to locate maximum bending moment).
•  Negative shear implies decreasing bending moment. Even though the forgoing is for uniform load, it applies to concentrated load and non-uniform load as well.  The derivation for such loads is similar.