Fig. 30 shows the arrangement of loading which causes a maximum bending moment at A in the beam. ΤΑ Fig. 31. Arrangement of Loading which Causes a Maximum Bending Moment in the Column Fig. 31 shows the arrangement of loading which causes a maximum bending moment at A in the column. CHAPTER VII BENDING AND DIRECT COMPRESSION 1. General Conditions. In the preceding formulas involving flexural stresses it was assumed that the external forces or their resultant acted perpendicularly to the axis of a member. In reinforced-concrete construction, however, members are frequently subjected simultaneously to both BENDINGSTRESSES and DIRECT AXIAL STRESSES. This is the condition, for example, in a beam subjected to the action of inclined forces along its length, or to additional forces, tension or compression, acting at its extremities. Similarly, in the design of arch-rings, it is necessary to provide for a combination of bending stresses and direct stresses. In building-construction the consideration of the stresses developed by this combination of forces is confined chiefly to columndesign, although the formulas and diagrams in this chapter are of general application and may be very usefully employed in the design of beams. In all cases the resultant stresses for which the design must be adequate is a combination of DIRECT AXIAL STRESSES, almost invariably compressive, with those due to FLEXURE. 2. Notation. The following notation is used: A A, A', area of cross-section; A = bt for rectangular sections; = area of cross-section of steel near face of member, least stressed in com pression; = area of cross-section of steel near face of member, most highly stressed in compression; At = area of cross-section of transformed section; M Ꭱ = = = = = = total area of cross-section of steel in symmetrically placed reinforcement; Ao As + A's; = moment of inertia of cross-section of concrete, about the gravity axis; moment of inertia of cross-section of steel, about the gravity axis; moment of inertia of transformed cross-section, about the gravity axis, I= I + nIs; resultant of all forces acting on or all stresses acting in the section; component of R normal to the section; a = distance from median axis of cross-section to steel, in sections with c = distance from median axis to face of section, in plain concrete; C1 = distance from gravity axis, to face most highly stressed in compression; c2= distance from gravity axis, to face least stressed in compression; C2 di d2 e= fs = = = distance from centroid of cross-section of steel, A's, to adjacent face of section; distance from centroid of cross-section of steel, A,, to adjacent face of section; eccentricity, or distance from median or gravity axis of cross-section to point of application of N; unit fiber-stress in steel near face of member, least stressed in compression; f', unit fiber-stress in steel near face of member, most highly stressed in fc f"c Ρ kt = = = = = compression; maximum compressive fiber-stress in concrete; minimum compressive fiber-stress in concrete; percentage of steel in sections with symmetrical reinforcement; Ao/bt; distance from the compressive surface to the neutral surface; r = radius of effective concrete section, in circular sections; t = total thickness or depth of cross-section of concrete. 3. Plain-Concrete Members. Referring to Fig. 1, if the line BC represents in projection a plain-concrete section, and R the RESULTANT of all forces acting on the section, then it is apparent that if R were applied at O, the center of gravity of the section, the unit stress would be UNIFORM Over the entire area A of the cross-section, and equal in intensity to N, the normal component Median Axis of R, divided by A, or Since, however, the result N B ant R is not applied at O, but at some other point Q, at a distance e from O, a moment, M, is produced about the point 0, the value of which is Ne. If e then represents the LEVER-ARM or the ECCENTRICITY of the component N, about the center of gravity 0, the Fig. 1. effect of the force N at Q is equivalent to the COM- Member. Stress-DiaBINED EFFECTS of an equal force N at O and a COUPLE, gram for Eccentric Load. the moment of which is M Therefore, in a plain-concrete section without reinforcement the TOTAL N UNIT COMPRESSIVE STRESS in the concrete due to an eccentric load N is = Ne. plus A the additional unit stress due to the eccentricity and determined by the flexureformula, M = fI/c. Or, the COMBINED STRESS IS For a rectangular cross-section of breadth b and thickness t, A I = bt3/12. Since the MAXIMUM COMPRESSIVE AND TENSILE STRESSES DUE TO = = bt and FLEXURE are at the most remote fibers, or at the edges of the section, c equals t/2. Also, as noted above, M = Ne. Substituting these values in the equation, The maximum stress is always compressive. The minimum stress is compressive when 6e/t is less than unity, for it is evident that if the eccentricity, e, is less than one sixth the thickness t, the resultant passes within the MIDDLE THIRD and there is no tension. If 6e/t is greater than unity, however, the +87 lb Median Axis- N=300000 16 3" 82 ----72*------72:-- - 24" +607 lb Fig. 2. Plain-Concrete Member. Stress-Diagram Showing Compression over Entire Section minimum stress is negative and there is tension over a part of the section. If the value of this unit tensile stress is greater than that which the plain concrete can safely resist, which should be not over 50 lb per sq in, the section must be increased, or reinforcement employed, in which latter case the procedure for reinforced-concrete members must be adopted. Example 1. A plain-concrete pier, 6 ft in height has a thickness t, 24 in, and a breadth b, 36 in. It is required to determine the MAXIMUM AND MINIMUM UNIT FIBER-STRESSES when the pier sustains a load N of 300 000 lb applied at a distance of e = 3 in from the median axis (Fig. 2). By Formula (2) As the eccentricity, e, is less than one sixth the thickness t, the resultant falls within the middle third, and the minimum unit stress determined by Formula (3), is compressive. |