Architectural Structures

Bending and Shear

Although derivation and numeric examples are required to analyze and design beams, intuitive understanding is an important prerequisite to gain deeper insight into the behavior of beams. The following is an intuitive introduction to beam bending and shear. A simple beam with uniform load is used for convenience.

Bending moment

Gravity load on a simple beam shortens the top and elongates the bottom, causing compressive and tensile stresses at top and bottom, respectively; with zero stress at the neutral axis (N. A.). In beams of symmetrical cross-section, the neutral axis is at the center. The compressive and tensile stress blocks generate an internal force couple that resists the external bending moment caused by load.

1Simple beam with pin and roller supports

2Deformed beam under uniform gravity load

3Free-body diagram with bending stress block that generates an internal force couple to resist the external bending moment caused by load

6-4 FUNDAMENTALS: Bending

Shear force

With few exceptions, described later, shear coexists with bending. When shear is present it acts both horizontally and vertically at equal magnitude. In wood beams horizontal shear is more critical because wood’s shear capacity is much smaller parallel than perpendicular to the grain.

1Beam under uniform load with shear cracks as they occur in some concrete beams near the supports where shear is maximum

2Tendency of beam parts to slide vertically generates vertical shear stress that is zero at mid-span and increases to maximum at the supports where the vertical shear deformation is greatest

3Tendency of beam layers to slide horizontally generates horizontal shear that is zero at mid-span and increases toward the supports. This is visualized, assuming a beam composed of several boards

4Shear diagram reflects shear distribution over beam length

5Unloaded beam marked with squares to visualize shear

6Loaded beam with squares deformed into rhomboids due to shear

7Horizontal and vertical shear couples on a square beam part are equal to balance rotational tendencies (ΣM = 0). Therefore, horizontal and vertical shear stresses are equal at any point on the beam.

8Shear vectors generate compression and tension diagonal to the shear. This tends to generate diagonal tension cracks in concrete beams

6-5 FUNDAMENTALS: Bending

Bending and shear stress

Bending 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.

1Simple wood beam with hatched area and square marked for inquiry

2Shear diagram with hatched area marked for inquiry

3Bending moment diagram with hatched area marked for inquiry

4Partial beam of length x, with stress blocks for bending fb and shear fv, where x is assumed a differential (very small) length

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) = fbbd2/6, or

fb= M/(bd2/6); where bd2/6 = S= Section Modulus for rectangular beam; thus fb= M/S

* 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. Let M = bending moment at x, and fb= bending stress at x, then M = RX = VX, and fb=M/S=Vx/S. Substituting Vx/S for fb in C = fbbd/4, the compressive top force, yields C=(Vx/S)(bd/4). Thus fv=C/(bx) yields fv=(Vx/S)(bd/4)/(bx). Substituting bd2/6 for S yields

fv = 1.5 V / bd

Equilibrium Method

Cantilever beam with point load

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 bendingand shear stress, similar to axial load yielding axial stress f= P/A. Formulas for bendingand shear stress are given on the next page and derived later in this chapter.

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 +)

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.

6-8 FUNDAMENTALS: Bending

Simple beam with uniform load

1Beam of L= 20 ft span, with uniformly distributed load w = 100 plf

2Free-body diagram of partial beam x units long

3Shear diagram

4Bending 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

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)L/2 – (wL/2)L/4 = 2wL2/8 - wL2/8, or

Mmax = wL2 / 8

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

Examples

The following wood beams demonstrate the area method for design and analysis. For design, a beam is sized for given loads; for analysis, stresses are checked against allowable limits, or how much load a beam can carry.

Beam design

V Shear diagram. M Bending diagram.

∆Deflection diagram.

I Inflection point (change from + to - bending).

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

6-11 FUNDAMENTALS: Bending

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.

6-12 FUNDAMENTALS: Bending