## Wednesday, April 30, 2014

### STRUCTURES SIMPLE BEAM WITH UNIFORM LOAD

1  Beam of L= 20 ft span, with uniformly distributed load w = 100 plf
2  Free-body diagram of partial beam x units long
3 Shear diagram
4  Bending moment diagram

To find the distribution of shear and bending along the beam, we investigate the beam at intervals of 5’, from left to right.  This is not normally required.

Reactions R are half the load on each support due to symmetry

Shear force Vx at any distance x from left is found using ∑ V = 0

V  = R - w x

Bending moment Mx at any distance x from left is found by ∑ M = 0.

∑ M = 0;    R x – w x (x/2) - Mx = 0;   solving for Mx

Bending is zero at both supports since pins and rollers have no moment resistance. Since the bending formula Mx= Rx-wx2/2 is quadratic, bending increase is quadratic (parabolic curve) toward maximum at center, and decreases to zero at the right support. For simple beams with uniform load the maximum shear force is at the supports and the maximum bending moment at mid-span (x= L/2) are:

Vmax = R = w L / 2

Mmax = wL^2 / 8

This formula is only for simple beams with uniform load.  Verifying example:

## Wednesday, April 23, 2014

### STRUCTURES EQUILIBRIUM METHOD

Assume a beam of length L = 10 ft, supporting a load P = 2 k.  The beam bending moment and shear force may be computed, like the external reactions, by equations of equilibrium ΣH=0, ΣV= 0, and ΣM=0.  Bending moment and shear force cause bending-and shear stress, similar to axial load yielding axial stress f= P/A.

1      Cantilever beam with concentrated load
V     Shear diagram (shear force at any point along beam)
M    Bending moment diagram (bending moment at any point along beam)
∆     Deflection diagram (exaggerated for clarity)

Reactions, found by equilibrium, ΣV=0 (up +) and ΣM=0 (clockwise +)

Shear V, found by vertical equilibrium, ΣV=0 (up +)

Left of a and right of b, shear is zero because there is no beam to resist it (reaction at b reduces shear to zero).  Shear may be checked, considering it starts and stops with zero. Concentrated loads or reactions change shear from left to right of them.  Without load between a and b (beam DL assumed negligible) shear is constant.

Bending moment M, found by moment equilibrium, ΣM=0 (clockwise +)

The mid-span moment being half the moment at b implies linear distribution.  The support reaction moment is equal and opposite to the beam moment.

Deflection ∆ is described later.  Diagrams visualize positive and negative bending by concave and convex curvature, respectively.  They are drawn, visualizing a highly flexible beam, and may be used to verify bending.

## Wednesday, April 16, 2014

### Structures: Bending and Shear Stress

ending and shear stresses in beams relate to bending moment and shear force similar to the way axial stress relates to axial force (f = P/A).  Bending and shear stresses are derived here for a rectangular beam of homogeneous material (beam of constant property).  A general derivation follows later with the Flexure Formula.

Simple wood beam with hatched area and square marked for inquiry
2  Shear diagram with hatched area marked for inquiry
3  Bending moment diagram with hatched area marked for inquiry
4  Partial beam of length x, with stress blocks for bending fb and shear fv, where x is assumed a differential (very small) length

Reactions, found by equilibrium ΣM = 0 (clockwise +)

Shear V, found by vertical equilibrium, ΣV=0 (upward +).

Bending moment M, found by equilibrium ΣM=0 (clockwise +)

Bending stress fb is derived, referring to 4.  Bending is resisted by the force couple C-T,  with lever arm 2/3 d = distance between centroids of triangular stress blocks.  C=T= fb bd/4, M= C(2d/3) = (fbbd/4)(2d/3) = fbbd^2/6, or  fb= M/(bd^2/6); where bd^2/6 = S= Section Modulus for rectangular beam; thus

* multiplying by 1000 converts kips to pounds, by 12 converts feet to inches. Shear stress fv is derived, referring to 4.  Bending stress blocks pushing and pulling in opposite directions create horizontal shear stress.  The maximum shear stress is fv=C/bx, where b = width and x = length of resisting shear plane.  Shear at left support is V = R.

## Thursday, April 3, 2014

### Structures: Bending Elements

Bending elements are very common in structures, most notably as beams.  Therefore, the theory of bending is also referred to as beam theory, not only because beams are the most common bending elements but their form is most convenient to derive and describe the theory.  For convenience, similar elements, such as joists and girders, are also considered beams.  Although they are different  in the order or hierarchy of structures,  their bending behavior is similar to that of beams, so is that of other bending elements, such as slabs, etc., shown on the next page. Thus, although the following description applies to the other bending elements, the beam analogy is used for convenience.

Beams are subject to load that acts usually perpendicular to the long axis but is carried in bending along the long axis to vertical supports.  Under gravity load beams are subject to bending moments that shorten the top in compression and elongate the bottom in tension. Most beams are also subject to shear, a sliding force, that acts both horizontally and vertically. Because beams and other bending elements are very common, the beam  theory is important in structural design and analysis.

As for other structural elements, beam investigation may involve analysis or design;  analysis, if a given beam is defined by architectural or other factors; design, if beam dimensions must be determined to support applied loads within allowable stress and deflection. Both, analysis and design, require to find the tributary load, reactions, shear, and bending moment.  In addition, analysis requires to find deflections, shear- and  bending stress, and verify if they meet allowable limits; by contrast design requires sizing  the beam, usually starting with an estimated size.

The following notations are commonly used for bending and shear stress:

Allowable stresses are given in building codes for various materials.

Allowable stresses assumed in this chapter are:

The more complex design and analysis of concrete and masonry will be introduces-later.