t2, 3, and may be neglected without introducing much error, and we get If the square tube be of uniform thickness, t, throughout d1 = d - 2 t. Substituting this value of d, in equation (12) and neglecting all terms containing t2, t3, and t4, we get— This equation gives a sufficiently accurate result if t be small compared with d. 140. Beams of H-Section with Equal Flanges. Substituting these values of I and h in equation (3), we get— If the thickness of the flanges be small compared with the depth of the beam, we obtain from equation (13), Chapter IX. If the web of the girder be thin, it may be neglected without introducing much error; in which case we find from equation (14), Chapter IX. M = ƒ do a (15). If the beam be placed so that its web is horizontal instead of vertical; its moment of inertia, with respect to a horizontal axis passing through its centre of gravity, is from equation (10), Chapter IX. where b, d, b1, d, represent the dimensions already given. From this we get b } 61 An adaptation of equation (15) may be used for determining the strengths of most wrought-iron and steel-rivetted girders, as in such cases the webs are thin, and for practical purposes the strength which they add to the girder in resisting bending moments may be neglected. In such girders, also, the thickness of the flanges is small compared with the depth of the girder. total stress on either flange of the girder in tons; s = ƒ av Let S then where f = = = unit-stress in either flange, а1 sectional area of either flange in square inches. Equation (15) may, therefore, be written where d depth between centres of gravity of the flanges. = (17), For girders resting on two abutments of span l and supporting For a cantilever of length, 7, with a weight, W, at the end, For a cantilever supporting a distributed weight, W, (20). Example 1.-A cast-iron girder of H-section rests on two supports 20 feet apart; what weight placed at its centre will break it, the modulus of rupture being 12 tons; and the section of the girder being Total depth d= 12 inches, Thickness of web b1 = inches, = 1 inch? The flanges are of equal thickness, viz., 1 inches, = Let W required breaking weight in tons. Substitute in equation (13) = 60 W. {b d3 − (b − b1) d ̧3} 6 x 12 {5 x (12)3-(5-1) x (9)3}. = 15.9 tons. Applying the approximate equation (14), we get 141. Beams of H-Section with Unequal Flanges. In cast-iron girders of H-section, the section of the bottom flange is made considerably larger than that of the top, as cast iron is much stronger in compression than in tension. The moment of resistance of the section of such a girder is expressed by the general equation where I == moment of inertia of the section with respect to an axis passing through its centre of gravity, and parallel to the flanges. = f unit-stress on the extreme fibres of the beam, h distance of extreme fibres from the neutral axis. = f may represent the unit-stress on the extreme fibres either at f the top or bottom of the beam, whichever gives to its least h value. For example, if the ultimate strength of cast iron in compression 40 tons per square inch, and the distance of the extreme top fibres from the neutral axis 8 inches, then = 8 In the same beam, if the strength of the iron in tension tons per square inch, and the distance of the extreme bottom in the general equation in determining the strength of the beam. width of top flange, thickness of top flange, b1 width of bottom flange, I = = = = = = = thickness of bottom flange, distance of top of beam from the neutral axis, unit-stresses at the top and bottom of the beam, 1 3 {b d3 − (b − t2) (d − t)3 + b ̧ d ̧3 − (b1 − tî) (d1 — t1)3}. See equation (g), page 150. Substituting this value of I in equation (2) we obtain {b d3 − (b − t) (d − t)3 + b1 d ̧3 − (b1 − t1⁄2) (d1 — t1)3} (22). fi 3d1 This is the complete expression for the moment of resistance of the section. Several approximations may be made according to the relative proportions of the different parts of the section. For example, if the thickness of the flanges be small compared with the depth of the beam, they may be supposed to be concentrated at their centre lines. In such case we get the following approximate formula distance of centre of top flange above the neutral axis, distance of centre of bottom flange below the neutral axis, = area of top flange, = area of bottom flange, area of web above the neutral axis, = area of web below the neutral axis, f = stress on the extreme fibres at the top of the beam, f stress on the extreme fibres at the bottom of the |