hapter 7 FORES IN BEAMS AN ABLES onsider a straight two-force member AB subjected at A and B to equal and opposite forces F and -F directed along AB. utting the member AB at and drawing the free-body B -F diagram of portion A, we conclude that the internal forces which existed at in member AB are equivalent to an axial force -F -F equal and opposite to F. A For a two-force member which is A not straight, the internal forces F reduce to a force-couple system and not to a single force. F
T y J x T M J F V A x A B F BE onsidering multiforce member A, cutting it at J, and drawing the freebody diagram of portion J, we conclude that the internal forces at J are equivalent to a force couple system consisting of the axial force F, the shearing force V, and a couple M. A y
T A x y B A J x F BE A y The magnitude of T the shearing force V measures the shear J at point J, and the M moment of the couple F is referred to as the bending moment at J. Since an equal and opposite force-couple system would have been obtained by considering the free-body diagram of portion AJ, it is necessary to specify which portion of member A was used when recording the answers.
Beams are usually long, straight prismatic members designed to support loads applied at various points along the member. In general, the loads are perpendicular to the axis of the beam and produce only shear and bending in the beam. The loads may be either concentrated at specific points, or distributed along the entire length or a portion of the beam. The beam itself may be supported in various ways. Since only statically determinate beams are considered in this text, we limit our analysis to that of simply supported beams, overhanging beams, and cantilevered beams.
To obtain the shear V and bending moment M at a given point of a beam, we first determine the reactions at the supports by considering the entire beam as a free body. We then cut the beam at and use the free-body diagram of one of the two portions obtained in this fashion to determine V and M.
M M V V The sign convention for positive shear force and bending moment is as shown. Once the values of shear and bending moment are established at several select points along the beam, it is usually possible to draw the shear diagram and bending-moment diagram for the entire beam.
M M V V onstructing these diagrams is generally facilitated by using the relationships dv dx = -w Integrating these, we find dm dx = V V -V = -(area under the load curve between and ) M -M = area under the shear curve between and
A y L A A x d B y 1 2 3 P 1 P 2 P3 B B x x 1 x 2 x 3 For a flexible cable with negligible weight supporting concentrated loads, and using the entire cable AB as a free body, the three available equations of equilibrium are not sufficient to determine the four unknown reactions at supports A and B.
A y L A x 1 1 A x P 1 P 2 P3 x2 d 2 3 B B y B x Knowing the coordinates of point, an additional equation can be obtained by constructing the freebody diagram of portions A or B of the cable. x 3 Once the reactions are known, the elevation of any point of the cable and the tension in any section of the cable can be determined.
T O θ T T θ W W T O For a cable carrying a distributed load we observe that the horizontal component of the tension T at is constant and must be equal to the tension T O at. The magnitude and direction of T are obtained from the force triangle. 2 T = T O + W 2 tan θ = W T O
A y (x,y) w B x In a suspension bridge, the load is uniformly distributed along the horizontal. The load supported by portion of the cable is W = wx. The curve formed by the cable is a parabola of equation y = wx2 2T o
A y O s c B (x,y) x For a cable hanging under its own weight, the load is uniformly distributed along the cable itself. The load supported by portion of the cable is W = ws. hoosing the origin O at a distance c = T o /w below, the relations between the geometry of the cantenary and cable tension are x x s = c sinh c y = c cosh c y 2 - s 2 = c 2 T O = wc W = ws T = wy