Bending moments shear force and torque

Consider the cantilever shown in Figure 2.1 (a), of length L metres and carrying a load of W kN at its extremity. The cantilever consists of thin top and bottom flanges, and of a web joining them together.

The load W creates a bending moment in the cantilever, which at a distance l from the end is Wxl and which reaches a maximum of WxL at the root of the cantilever, Figure 2.1 (b). The bottom flange of the cantilever is stressed in compression, and the top flange is stressed in tension, Figure 2.1 (c). The top and bottom surfaces of the cantilever are called the top and bottom extreme fibres, or the extrados and the intrados, respectively. The bending moment is designated as hogging, as it causes tension on the top fibre. Bending moments that cause tension on the bottom fibre, for instance due to an upwards load on the end of the cantilever, are designated as sagging. The distance between the centres of the compression and tension flanges is h. The compression and tension forces in the flanges, ±F, create an internal couple which must balance the external applied moment (ignoring very small longitudinal forces in the thin web). Consequently at the root of the cantilever, Fh = WL. As stress = force/area, the stress in each flange o = F/A, where A is the cross-section area of each flange.

The cantilever tip deflects downwards as the top flange extends and the bottom flange compresses. The amount of deflection depends on the height and thickness of the web, on the width and thickness of the flanges and on the stiffness of the material of which the cantilever is made.

The deflection 5 = WL3/3EI where

W is the load on the cantilever end; L is the length of the cantilever;

I is the 'moment of inertia' of the cross section of the cantilever, and is a measure of the strength given by its geometry;

E is the Young's modulus (also called the 'modulus of elasticity' or just the 'modulus') of the constituent material of the cantilever, and is a measure of the stiffness of the material. For instance, concrete has a modulus of, typically, 30,000 MPa, while steel, which is much stiffer, has a modulus of 200,000 MPa.

If the load W was applied suddenly, the cantilever would vibrate as well as deflecting.

There are two ways in which the cantilever may collapse. The first is in bending; either the top or bottom flanges may not be strong enough, when they would fail by extension or crushing, respectively, Figure 2.1 (d). The other is if the web joining them is not strong enough to hold the two flanges together, when the failure would be in shear, Figure 2.1 (e).

Shear force is always proportional to the slope of the bending moment diagram, shown in Figure 2.1 (b). As here this slope is constant, the shear force is also constant, Figure 2.1 (f), and equal to the load W. The action of shear is best represented by an analogy, in which the web is considered to be an 'N' truss, whose diagonal members

Figure 2.1 Bending moment and shear force on cantilever

are stressed in compression and the vertical members in tension, Figure 2.1 (g). This diagram also shows the conventional representation of tension and compression forces in members which is used in this book. Truss analogy is explained in more detail in 3.10.

The cantilever could have consisted of a beam of rectangular cross section, Figure 2.2 of width b and height h. The tensile and compressive stresses caused by the bending moment are zero at what is termed the neutral axis, which is at the centre of the beam for a symmetrical cross section, and they are proportional to their distance from this neutral axis. Consequently they are a maximum at the top and bottom extreme fibres where they are represented by the symbol ±a. As force = area X stress, the forces of tension and compression forming the internal couple are equal to the average stress in the top or bottom half of the beam multiplied by the cross-section area of half the beam, F = ±(o/2)x(bh/2) = ±obh/4. The lever arm of the internal couple is the distance between the centroids of the tension and compression forces. As the force diagrams are triangular, this lever arm is 2h/3. The stress on the top and bottom extreme fibres of the cross section created by the applied moment WL can be found by equating the external moment with the internal couple, WL = Fx2h/3. Substituting for F, WL = (obh/4)x2h/3, or WL = obh2/6, or a = WL/(bh2/6). The term bh2/6 is known as the elastic modulus of the rectangular cross section (not to be confused with E, the modulus of elasticity).

For bridge decks with cross sections that are unsymmetrical about a horizontal axis, Figure 2.3, the elastic moduli corresponding to the top and bottom extreme fibres are not equal. They are conventionally designated by zt and zb for the top and bottom extreme fibres respectively. Thus the stresses on the top and bottom extreme fibres of the bridge deck, subjected to an external bending moment M are M/zt and M/zb.

Figure 2.2 Rectangular cross section cantilever
Figure 2.3 Section unsymmetrical about a horizontal axis

Figure 2.4 Eccentric load creating torsion

If the load W had been applied to the cantilever off centre, Figure 2.4, in addition to the bending moment and the shear force that remain unchanged, the beam would be twisted. If the eccentricity of the load is d, the twisting moment, or torque, is Wxd.

2.5 Limit states

In the past, structures were designed to respect limiting stresses. In the cantilever described above, the tensile and compressive stresses at the top and bottom of the cantilever would have to remain below values that were considered safe for the material used in its construction. This would govern the maximum load W that the cantilever may safely carry.

The more enlightened modern tendency is to design to respect 'limit states'. Structures must perform satisfactorily in normal service, and must also have a sufficient margin of safety against collapse. The normal service condition is called the Serviceability Limit State (SLS). The SLS includes criteria on the deflection of structures when this may affect performance, such as damaging floor finishes or interfering with the drainage of rainwater, on their susceptibility to vibrate excessively and, critically, on their durability. Reinforced concrete deteriorates principally by the corrosion of the steel reinforcement within it. The useful life of a reinforced concrete member is controlled by the thickness and the quality of the protective concrete layer outside the reinforcement, called the cover, and by the width of the cracks in the concrete that occur in service (3.5, 3.6).

The collapse condition is called the Ultimate Limit State (ULS). The structure must not collapse when the loads applied to it are factored up by Load Factors, and the strength of its constituent materials is factored down by Material Factors. The magnitude of the Load Factors depends principally on the uncertainty that exists over the likely intensity of any load. For instance, the self weight of a bridge deck may vary only within narrow limits, due to variations in the density of materials and to tolerance on the size of the members, and as a result the Load Factor applied to self weight is low, typically 1.15. On the other hand, the traffic loading on a bridge deck is inherently variable and there is the possibility that under exceptional circumstances it may be heavier than expected, leading to a higher Load Factor, typically 1.3 - 1.5.

The Material Factor on steel reinforcement is low, typically 1.05 - 1.15, as steel is produced by an industrial process and there is little variation in its strength. On the other hand concrete is mixed on site and is inherently a variable product, leading to the need for a higher Material Factor, typically 1.5.

2.6 Statical determinacy and indeterminacy

The concepts of statical determinacy and indeterminacy re-occur in the text. A short explanation, or reminder, is given below.

Structures are determinate if the forces applied on their supports, called the support reactions, can be calculated using the two basic equations of equilibrium:

• the moments about any point sum to zero; Equation 1

• the forces in any direction sum to zero. Equation 2

Consider the statically determinate beam shown in Figure 2.5 (a), with a span L carrying two loads W1 and W2 which are situated respectively at distances l1 and l2 from point A. The beam is assumed to rest on bearings that allow rotation, the reactions at the bearings being R1 and R2. Take moments about point A:

Equation 1 W1xl1 + W2xl2 - R2xL = 0, which gives R2 = (W1xl1 + W2xl2)/L.

Resolve the forces and reactions in the vertical plane:

Equation 2 R1 + R2 - W1 - W2 = 0, which gives R1 = W1 + W2 - R2.

As the loads and their positions on the span are known, the equations may be simply solved to yield the value of the reactions. Once the reactions are known, the bending moments and shear forces at any point of the beam can be calculated.

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Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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