CHAPTER VIII. BEAMS. In the previous chapter the case of a beam loaded at different points over a certain space was considered. The object of referring to a beam was to apply the principle of moments to find what we termed the reactions on the supports. But the loads on this beam and any other beam set up stresses in the beam itself, and it is these which will be investigated. In the present chapter any piece of a structure lying across two supports, whether loaded or not, is called a beam. A floor joist is a beam, likewise a girder. A lattice girder of a railway bridge is a trussed beam. The tubes of the Britannia Bridge are all beams, although, of course, of a very elaborate kind; in fact, large built-up steel beams are usually called girders. When we load a beam it undergoes what is called a bending moment. In order that the beam may be safe there must also be a resisting moment equal to the bending moment. It is the material of the beam which supplies the resisting moment (to breaking), which, according to the factor of safety employed, may be five or six times, etc., the bending moment. In the sequel, BM stands for bending moment and RM for resisting moment. Again, we may have a beam not supported at both ends but supported, and fixed at one end only; it is then called a cantilever (or semi-beam). Beams may also be fixed at both ends. The various calculations in them then become different. Now the bending moment of a beam at any section is measured by the moment about that section of the applied forces tending to turn the portion of the beam on the right or left hand of the section. For instance, in fig. 101 we have a beam over a span of 12 ft. carrying 10 tons at the centre. The BM at the middle will be found by considering the forces on the right or left-hand side. We obviously have 2 equal reactions = 30 ton-feet. That is the bending moment. of 5 tons each .. 5 x 12 2 = Again, if we 9)-(10 × 3), .. (5 × 9) But we may want the bending moment at a point A, a point 3 ft. from the left-hand support. Here we have a BM of 5 x 3 = 15 ton-ft. consider the right-hand portion, we have two quantities, viz. (5 x the force of 5 tons being upwards and 10 tons downwards. (10 × 3) = 45-3015 ton-ft. as obtained before. In fact, to get the BM of any beam 1. Calculate the reactions of the supports. 2. Calculate the algebraic sum of the moments about the section of all the force acting, (a) either on the right-hand portion, (b) or left hand portion of the beam. Now, as well as a bending moment in a beam, we have what is called a shearing force,—that is, the tendency of one portion to slide over the other. To calculate the total shearing force at any section of a beam, take the algebraic sum of all the applied forces acting either (a) on the right-hand side of the beam, (b) on the left-hand side of the beam. For instance, in fig. 102 we have a beam loaded as shown. We want to find the shear at point C. Find the reaction on each support in the usual way their values are P 4.25 tons. I Q=4.75 tons. = Therefore on the left-hand portion we have an upward force (the reaction) of 4.25 and a downward one of 2 tons. .. shear 4.25 − 2 = 2·25 tons. Again, reaction at Q=4.75, and we have a downward force in this portion of 7 tons... 4·75 – 7 = 2.25 tons shear same as before. This is what is called a positive shear, because the preponderance of load on the right-hand side tends to cause that portion to slide down over the other in a clockwise direction. If the reverse be the case, there would be a negative shear. In any beam uniformly or centrally loaded we have the greatest BM at But the centre and the greatest shear at the supports, and nil at the centre. in a cantilever the BM is greatest at the supports, for this reason that we find in built-up girders more web stiffeners at the ends and more boom plates at the centre, because the booms take the BM and the webs or lattice work the shear. Any observant engineering pupil will have seen this. Beams are made of various sections, rectangular and I beams being the most common. The latter are usually steel; they were some years ago frequently cast iron. Now the strength of any beam is proportional to 1. Breadth, depth squared, 2. Inversely as the span, B and D measured in inches and L in feet. Taking the two most common forms of beams, we have for one of I section If Fallowable tensile stress in lbs. per sq. in., for wrought iron 5 tons for cast iron 1 tons, and steel 6 tons. = If A = area of top flange in square inches, and D depth (in.), ther the resisting moment of that beam is and of a rectangular beam Fxaxd Fxbxd2 (55 (56 For instance, we have an 18" x 7" I beam. What is its resisting moment Assume the top flange to be 7.5 sq. in. in sectional area, and A, which is what is called the effective depth, and measured from the centre of the flanges, = 17", and F, 5 tons. Then 5 x 7.5 x 17 6375 in.-tons resisting moment. If then we have a BM due to load of 100 in.-tons, the beam will stand a load of 6.375 tons. = In tables of properties of I beams, however, we often see what is called the modulus of section. This is the number by which the safe stress must be multiplied to give us the RM. In the case of the 18" x 7" beam the section modulus is 127-6; this multiplied by the 5 tons safe stress = 638'0 only, 5 difference. For a beam of circular section (not usually employed) we have a RM of R = radius of beam in inches. FTR3 (57) We will have observed that makers of I beams usually make the flanges equal in area this is because the steel at present on the market is assumed to be equal in tensile to compressive strength. Now cast iron has only a tensile strength of about that in compression. Therefore any observer of cast-iron beams will observe them to have a large bottom flange, which always takes the tension, and a small top flange for the compression. In the case of cast-iron cantilevers the opposite occurs, as the top flange then takes all the tension and the bottom one the compression. Casting beams so in order to save metal has been deemed a bad practice, owing to contraction in cooling taking place unequally. This has caused many CI beams to fail, and is one (together with the fact that standard-sized steel joists are cheaper) of the reasons for their disuse. We shall now discuss the various formulæ in reference to beams which occur in practice. Case II. Cantilever variously loaded (fig. 104). Bending moment at support = (w1X1) + (W2X2) + (W3X3) Maximum shear (occurs at support) = w1+w2 + Wz Shear at any point = No. of weights behind W, viz. towards the end Case III. Uniformly loaded cantilever (fig. 105). W = weight per foot-run W = total load = WL Bending moment at end = {wL2 (60) wx2 (61) 2 Case IV. Combine Cases I. and III. treating separately (fig. 106). Case VII. Beam supported and loaded anywhere (fig. 109).—Max. BM occurs at point of W, viz. just below it. Its value is Case VIII. Beam supported and loaded (fig. 110).-With a number of loads in various positions, first of all find the reactions on the supports R1 and R2 (vide Chap. VII.). Then BM at any point Y = Ry - W3[y - (L x3)] . Case IX. Uniformly loaded beam (supported) (fig. 111). W total load = Shear at any point x is positive when x is greater than (65) L and negative 2 Case X. (fig. 112). First find R2 = (W1 × CA) + (W2 × DA) BA (66) Then to find any BM between A and B we take R2 and multiply by distance from R2 to the point in question.† Case XI. same as Case X. (fig. 113).-Take the total load W as concentrated at x, neglecting W, * L = distance between supports, as in fig. 109. + R2 reaction at A. |