Engineering Materials Design Lecture.6 the design of beams By Dr. Mohammed Ramidh
6.1 INTRODUCTION Finding the shear forces and bending moments is an essential step in the design of any beam. we usually need to know not only the maximum values of these quantities, but also the manner in which they vary along the axis. Once the shear forces and bending moments are known, we can find the stresses, strains, and deflections, 6.2 TYPES OF BEAMS, LOADS, AND REACTIONS Beams are usually described by the manner in which they are supported. For instance, a beam with a pin support at one end and a roller support at the other (Fig. 6-1a) is called a simply supported beam or a simple beam. FIG. 6-1 Types of beams: (a) simple beam, (b) cantilever beam, and (c) beam with an overhang The beam shown in Fig. 6-1b, which is fixed at one end and free at the other, is called a cantilever beam. The third example in the figure is a beam with an overhang (Fig. 6-1c). This beam is simply supported at points A and B (that is, it has a pin support at A and a roller support at B) but it also projects beyond the support at B. The overhanging segment BC is similar to a cantilever beam except that the beam axis may rotate at point B. Types of Loads Several types of loads that act on beams are illustrated in( Fig. 6-1. )When a load is applied over a very small area it may be idealized as a concentrated load, which is a single force. Examples are the loads P1, P2, P3, and P4 in the figure. When a load is spread along the axis of a beam, it is represented as a distributed load, such as the load q in part (a) of the figure. Distributed loads are measured by their intensity, which is expressed in units of force per unit distance (for example, newtons per meter or pounds per foot). A uniformly distributed load, or uniform load, has constant intensity q per unit distance (Fig. 6-1a). Avarying load has an intensity that changes with distance along the axis; for instance, the linearly varying load of( Fig.6-1b )has an intensity that varies linearly from q1 to q2. Another kind of load is a couple, illustrated by the couple of moment M1 acting on the overhanging beam (Fig. 6-1c).
Reactions: Finding the reactions is usually the first step in the analysis of a beam. Once the reactions are known, the shear forces and bending moments can be found, an additional equation of equilibrium is then available for use in solving for the unknown support reactions included in that FBD. As an example, let us determine the reactions of the simple beam AB of ( Fig. a.) This beam is loaded by an inclined force P1, a vertical force P2, and a uniformly distributed load of intensity q. We begin by noting that the beam has three unknown reactions: a horizontal force HA at the pin support, a vertical force RA at the pin support, and a vertical force RB at the roller support The equation of horizontal equilibrium is from which we get, To find the vertical reactions RA and RB we write equations of moment equilibrium about points B and A, respectively, with counterclockwise moments being positive: As a second example, consider the cantilever beam of( Fig. b.) The loads consist of an inclined force P3 and a linearly varying distributed load. The latter is represented by a trapezoidal diagram of load intensity that varies from q1 to q2. The reactions at the fixed support are a horizontal force HA, a vertical force RA, and a couple MA. Equilibrium of forces in the horizontal direction gives and equilibrium in the vertical direction gives
The moment reaction MA at the fixed support is found from an equation of equilibrium of moments. In this example we will sum moments about point A in order to eliminate both HA and RA from the moment equation. Also, for the purpose of finding the moment of the distributed load, we will divide the trapezoid into two triangles, as shown by the dashed line in( Fig. b.) Each load triangle can be replaced by its resultant, which is a force having its magnitude equal to the area of the triangle and having its line of action through the centroid of the triangle. Thus, the moment about point A of the lower triangular part of the load is in which q1b/2 is the resultant force (equal to the area of the triangular load diagram) and L 2b/3 is the moment arm (about point A) of the resultant. The moment of the upper triangular portion of the load is obtained by a similar procedure, and the final equation of moment equilibrium (counterclockwise is positive) is The beam with an overhang (Fig. c) supports a vertical force P4 and a couple of moment M1. Since there are no horizontal forces acting on the beam,( Let us arbitrarily decide to write two moment equations, the first for moments about point B and the second for moments about point A,
6.3 SHEAR FORCES AND BENDING MOMENTS When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam. To determine these stresses and strains, we first must find the internal forces and internal couples that act on cross sections of the beam. As an illustration of how these internal quantities are found, consider a cantilever beam AB loaded by a force P at its free end (Fig.a). We cut through the beam at a cross section mn located at distance x from the free end and isolate the lefthand part of the beam as a free body (Fig. b). The action of these same stress resultants against the right-hand part of the beam is shown in (Fig. c.) The directions of both quantities are now reversed the shear force acts upward and the bending moment acts clockwise. However, the shear force still tends to rotate the material clockwise and the bending moment still tends to compress the upper part of the beam and elongate the lower part. 6.4 RELATIONSHIPS BETWEEN LOADS, SHEAR FORCES, AND BENDING MOMENTS We will now obtain some important relationships between loads, shear forces, and bending moments in beams. These relationships are quite useful when investigating the shear forces and bending moments throughout the entire length of a beam, and they are especially helpful when constructing shear-force and bending-moment diagrams. As a means of obtaining the relationships, let us consider an element of a beam cut out between two cross sections that are distance dx apart (Fig. ). The load acting on the top surface of the element may be a distributed load, a concentrated load, or a couple, as shown in Figs. a, b, and c, respectively Distributed Loads (Fig. a) Shear Force. Equilibrium of forces in the vertical direction (upward forces are positive) gives..(6-1)
Bending Moment. Let us now consider the moment equilibrium of the beam element shown in (Fig. a.).(6-2) Concentrated Loads (Fig. 4-14b) Now let us consider a concentrated load P acting on the beam element (Fig. b). From equilibrium of forces in the vertical direction, we get From equilibrium of moments about the left-hand face of the element (Fig. b), we get.(6-3) Loads in the Form of Couples (Fig. c) The last case to be considered is a load in the form of a couple M0 (Fig. c). From equilibrium of the element in the vertical direction we obtain V1 = 0, which shows that the shear force does not change at the point of application of a couple. Equilibrium of moments about the left-hand side of the element gives..(6-4) 6.5 SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS When designing a beam, we usually need to know how the shear forces and bending moments vary throughout the length of the beam. of special importance are the maximum and minimum values of these quantities. Information of this kind is usually provided by graphs in which the shear force and bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as the abscissa. Such graphs are called shear-force and bending-moment diagrams.
Concentrated Load Let us begin with a simple beam AB supporting a concentrated load P (Fig. a). The load P acts at distance a from the left-hand support and distance b from the right-hand support. Considering the entire beam as a free body, we can readily determine the reactions of the beam from equilibrium; the results are (6-5) We now cut through the beam at a cross section to the left of the load P and at distance x from the support at A. Then we draw a free-body diagram of the left-hand part of the beam (Fig. b). From the equations of equilibrium for this free body, we obtain the shear force V and bending moment M at distance x from the support: (6-6) Next, we cut through the beam to the right of the load P (that is, in the region a x L) and again draw a free-body diagram of the left-hand part of the beam (Fig. c). From the equations of equilibrium for this free body, we obtain the following expressions for the shear force and bending moment..(6-7a)..(6-7b) Note that these equations are valid only for the right-hand part of the beam. The equations for the shear forces and bending moments (Eqs. 6-6 and 6-7) are plotted below the sketches of the beam.( Figure d) is the shear-force diagram and (Fig. e) is the bending-moment diagram. From the first diagram we see that the shear force at end A of the beam (x = 0) is equal to the reaction RA. Then it remains constant to the point of application of the load P. At that point, the shear force decreases abruptly
by an amount equal to the load P. In the right-hand part of the beam, the shear force is again constant but equal numerically to the reaction at B. As shown in the second diagram, the bending moment in the left hand part of the beam increases linearly from zero at the support to Pab/L at the concentrated load (x =a). In the right-hand part, the bending moment is again a linear function of x, varying from Pab/L at x = a to zero at the support (x =L). Thus, the maximum bending moment is..(6-8) Uniform Load Asimple beam with a uniformly distributed load of constant intensity q is shown in ( Fig. a) on the next page. Because the beam and its loading are symmetric, we see immediately that each of the reactions (RA and RB) is equal to ql/ 2. Therefore, the shear force and bending moment at distance x from the left-hand end are (6-9).(6-10) These equations, which are valid throughout the length of the beam, are plotted as shear-force and bending moment diagrams in( Figs. b and c,) respectively. The shear-force diagram consists of an inclined straight line having ordinates at x = 0 and x = L equal numerically to the reactions. The slope of the line is _q, as expected from Eq. (6-1). The bending-moment diagram is a parabolic curve that is symmetric about the midpoint of the beam. At each cross section the slope of the bendingmoment diagram is equal to the shear force (see Eq. 6-2): The maximum value of the bending moment occurs at the midpoint of the beam where both dm/dx and the shear force V are equal to zero. Therefore, we substitute x =L/ 2 into the expression for M and obtain.(6-11)
Example: Draw the shear-force and bending-moment diagrams for a cantilever beam with two concentrated loads (Fig. a). Solution Reactions. From the free-body diagram of the entire beam we find the vertical reaction RB (positive when upward) and the moment reaction MB Shear forces and bending moments. We obtain the shear forces and bending moments by cutting through the beam in each of the two segments, drawing the corresponding free-body diagrams, and solving the equations of equilibrium. Again measuring the distance x from the left-hand end of the beam, we get The corresponding shear-force and bending-moment diagrams are shown in Figs. b and c. Homework: 1- Determine the shear force V and bending moment M at the midpoint C of the simple beam AB shown in the figure. 2- The cantilever beam AB shown in the figure is subjected to a concentrated load P at the midpoint and a counterclockwise couple of moment M1 = PL/4 at the free end. Draw the shear-force and bending-moment diagrams for this beam. 3- Determine the shear force V and bending moment M at a cross section located 18 ft from the left-hand end A of the beam with an overhang shown in the figure
4. The simple beam AB shown in the figure is subjected to a concentrated load P and a clockwise couple M1 =PL/3 acting at the third points. Draw the shear-force and bending-moment diagrams for this beam. 5. Beam ABCD is simply supported at B and C and has overhangs at each end (see figure). The span length is L and each overhang has length L/ 3. A uniform load of intensity q acts along the entire length of the beam. Draw the shear-force and bending-moment diagrams for this beam. 6. A cantilever beam AB supports a couple and a concentrated load, as shown in the figure. Draw the shear force and bending-moment diagrams for this beam.
Stresses in Beams we saw how the loads acting on a beam create internal actions (or stress resultants) in the form of shear forces and bending moments. we go one step further and investigate the stresses and strains associated with those shear forces and bending moments. Knowing the stresses and strains, we will be able to analyze and design beams subjected to a variety of loading conditions. The deflection of the beam at any point along its axis is the displacement of that point from its original position, measured in the y direction. We denote the deflection by the letter v to distinguish it from the coordinate y itself (see Fig. 5-1b).* FIG. 6-3 Bending of a cantilever beam: (a) beam with load, and (b) deflection curve 6.6 DESIGN OF BEAMS FOR BENDING STRESSES The process of designing a beam requires that many factors be considered, including the type of structure (airplane, automobile, bridge, building, or whatever), the materials to be used, the loads to be supported, the environmental conditions to be encountered, and the costs to be paid. However, from the standpoint of strength, to selecting a shape and size of beam such that the actual stresses in the beam do not exceed the allowable stresses for the material. In this section, we will consider only the bending stresses (that is, the stresses obtained from the flexure formula, Eq This equation, called the flexure formula, shows that the stresses are directly proportional to the bending moment M and inversely proportional to the moment of inertia I of the cross section. Stresses calculated from the flexure formula are called bending stresses or flexural stresses. When designing a beam to resist bending stresses, we usually begin by calculating the required section modulus. if the beam has the allowable stresses are the same for both tension and compression, we can calculate the required modulus by dividing the maximum bending moment by the allowable bending stress for the material. (6-1)
If the cross section is not doubly symmetric, or if the allowable stresses are different for tension and compression, we usually need to determine two required section moduli one based upon tension and the other based upon compression. available shapes include wide-flange beams, I-beams, angles, channels, rectangular beams, and tubes. Beams of Standardized Shapes and Sizes Structural-steel sections are given a designation such as W 30 211 in USCS units, which means that the section is of W shape (also called a wide-flange shape) with a nominal depth of 30 in. and a weight of 211 lb per ft of length (see Table E-1(a), Appendix E). The corresponding properties for each W shape are also given in SI units in Table E-1(b). For example, in SI units, the W 30 211 is listed as W 760 314 with a nominal depth of 760 millimeters and mass of 314 kilograms per meter of length. Similar designations are used for S shapes (also called I-beams) and C shapes (also called channels), as shown in Tables E-2(a) and E-3(a) in USCS units and in Tables E- 2(b) and E-3(b) in SI units. Angle sections, or L shapes, are designated by the lengths of the two legs and the thickness (see Tables E-4 and E-5). For example, L 8 6 1 (see Table E-5(a)) denotes an angle with unequal legs, one of length 8 in. and the other of length 6 in., with a thickness of 1 in. The corresponding label in SI units for this unequal leg angle is L 203 152 25.4 (see Table E-5(b)). Relative Efficiency of Various Beam Shapes As an illustration, consider a cross section in the form of a rectangle of width b and height h (Fig. 6-4a). The section modulus (from Eq. ) is..(6-2) where A denotes the cross-sectional area. This equation shows that a rectangular cross section of given area becomes more efficient as the height h is increased (and the width b is decreased to keep the area constant). of course, there is a practical limit to the increase in height, Next, let us compare a solid circular cross section of diameter d (Fig. 6-4b) with a square cross section of the same area. The side h of a square having the same area as the circle is h = (d/2). The corresponding section moduli (from Eqs. S = S = ) are and FIG. 6-4 Cross-sectional shapes of beams
(6-3a) (6-3b) from which we get,.(6-4) This result shows that a beam of square cross section is more efficient in resisting bending than is a circular beam of the same area. The ideal cross-sectional shape for a beam as shown in Fig. 6-4c. For this ideal shape, we obtain..(6-5) (Fig. 6-4d). For standard wide-flange beams, the section modulus is..(6-6) Example 6-1 A simply supported wood beam having a span length L = 12 ft carries a uniform load q =420 lb/ft. The allowable bending stress is 1800 psi, the wood weighs 35 lb/, and the beam is supported laterally against sideways buckling and tipping. Select a suitable size for the beam from the table in Appendix F. Solution Since we do not know in advance how much the beam weighs, we will proceed by trial-and-error as follows: (1) Calculate the required section modulus based upon the given uniform load. (2) Select a trial size for the beam. (3) Add the weight of the beam to the uniform load and calculate a new required section modulus. (4) Check to see that the selected beam is still satisfactory. If it is not,select a larger beam and repeat the process. (1) The maximum bending moment in the beam occurs at the midpoint (see Eq., The required section modulus (Eq. 6-1) is
(3) The uniform load on the beam now becomes 426.8 lb/ft, and the corresponding required section modulus is Example 6-2 Avertical post 2.5-meters high must support a lateral load P =12 kn at its upper end. Two plans are proposed a solid wood post and a hollow aluminum tube. (a) What is the minimum required diameter d1 of the wood post if the allowable bending stress in the wood is 15 MPa? (b) What is the minimum required outer diameter d2 of the aluminum tube if its wall thickness is to be one-eighth of the outer diameter and the allowable bending stress in the aluminum is 50 MPa? FIG. 6-5 Example 5-6. (a) Solid wood post, and (b) aluminum tube
Solution Maximum bending moment. The maximum moment occurs at the base of the post and is equal to the load P times the height h; thus, Mmax = Ph =(12 kn)(2.5 m) =30kN.m (a) Wood post. The required section modulus S1 for the wood post (see Eqs. and 6-1) is Solving for the diameter, we get d1=273 mm (b) Aluminum tube. To determine the section modulus S2 for the tube, we first must find the moment of inertia I2 of the cross section. The wall thickness of the tube is d2/8, and therefore the inner diameter is d2 - d2/4, or 0.75d2. Thus, the moment of inertia (see Eq. I= ) is The section modulus of the tube is now obtained from Eq. (s = ) as follows: The required section modulus is obtained from Eq. (6-1): By equating the two preceding expressions for the section modulus, we can solve for the required outer diameter: The corresponding inner diameter is 0.75(208 mm), or 156 mm. Example 6-3 A temporary wood dam is constructed of horizontal planks A supported by vertical wood posts B that are sunk into the ground so that they act as cantilever beams (Fig. 6-6). The posts are of square cross section (dimensions b b) and spaced at distance S = 0.8 m, center to center. Assume that the water level behind the dam is at its full height h = 2.0 m. Determine the minimum required dimension b of the posts if the allowable bending stress in the wood is = 8.0 MPa.
FIG. 6-3 Example 6-6. Wood dam with horizontal planks A supported by vertical posts B Solution Loading diagram. Each post is subjected to a triangularly distributed load produced by the water pressure acting against the planks. Consequently, the loading diagram for each post is triangular (Fig. 6-6c). The maximum intensity of the load on the posts is equal to the water pressure at depth h times the spacing s of the posts:.(a) in which is the specific weight of water. Note that q0 has units of force per unit distance, has units of force per unit volume, and both h and S have units of length. Section modulus. Since each post is a cantilever beam, the maximum bending moment occurs at the base and is given by the following expression: Therefore, the required section modulus (Eq. 6-1) is For a beam of square cross section, the section modulus is S = /6 (see Eq. S = /6). Substituting this expression for S into Eq. (c), we get a formula for the cube of the minimum dimension b of the posts: Numerical values. We now substitute numerical values into Eq. (d) and obtain from which b =199 mm Thus, the minimum required dimension b of the posts is 199 mm. Any larger dimension, such as 200 mm, will ensure that the actual bending stress is less than the allowable stress.
Homework: 5.6-11 A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam AB (see figure). The load transmitted to the beam from the front axle is 2200 lb and from the rear axle is 3800 lb. The weight of the beam itself may be disregarded. (a) Determine the minimum required section modulus S for the beam if the allowable bending stress is 17.0 ksi, the length of the beam is 18 ft, and the wheelbase of the carriage is 5 ft. (b) Select the most economical I-beam (S shape) from Table E-2(a), Appendix E. 5.6-16 A beam having a cross section in the form of a channel (see figure) is subjected to a bending moment acting about the z axis. Calculate the thickness t of the channel in order that the bending stresses at the top and bottom of the beam will be in the ratio 7:3, respectively. 5.6-15 A beam having a cross section in the form of an unsymmetric wide-flange shape (see figure) is subjected to a negative bending moment acting about the z axis. Determine the width b of the top flange in order that the stresses at the top and bottom of the beam will be in the ratio 4:3, respectively. 5.6-23 The cross section of a rectangular beam having width b and height h is shown in part (a) of the figure. For reasons unknown to the beam designer, it is planned to add structural projections of width b/9 and height d to the top and bottom of the beam [see part (b) of the figure]. For what values of d is the bending-moment capacity of the beam increased? For what values is it decreased?