- Beams are structural member supporting lateral loadings, i.e., these applied perpendicular to the axes.

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Willis Barnett
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1 4. Shear and Moment functions - Beams are structural member supporting lateral loadings, i.e., these applied perpendicular to the aes. - The design of such members requires a detailed knowledge of the variations of the internal shear force V and moment M acting at each point along the ais of the beam. - The internal normal force is generally not considered for reasons. () In most cases the loads applied to a beam act perpendicular to the beam s ais and hence produce only an internal shear force and bending moment. () For design purposes the beam s resistance to shear, and particularly to bending, is more important than the ability to resist normal force - However when beams are subjected to compressive aial forces with normal loads, it can be important. The variation of V and M are a function of the position of an arbitrary point along the beam s ais can be obtained by using method of section. Procedure () Support reactions () Shear and moment function In general, the internal shear & moment functions will be discontinuous or their slope will be discontinuous at points where: - The type or magnitude of the distributed load changes - Concentrated forces or couple moments are applied - Because of this, shear and moment functions must be determined for each region of the beam located between any two discontinuities of loading. For eample, coordinates,, and will have to be used to describe the variation of V and M throughout the length of the beam in Fig. 4 5a. These coordinates will be valid only within regions from A to B for, from B to C for, and from C to D for. Although each of these coordinates has the same origin, as noted here, this does not have to be the case. Ref) Indeed, it may be easier to develop the shear and moment functions using coordinates,, having origins at A, B, and D as shown in Fig. 4 5b. Here and are positive to the right and is positive to the left. Procedure for Analysis Determine the support reactions on the beam Resolve all the eternal forces into components acting perpendicular & parallel to beam s ais Specify separate coordinates and associated origins, etending into: Regions of the beam between concentrated forces and/or couple moments; or Discontinuity of distributed loading -Section the beam perpendicular to its ais at each distance -From the free-body diagram of one of the segments, determine the unknowns V & M -On the free-body diagram, V & M should be shown acting in their + directions -V is obtained from -M is obtained by -The results can be checked by noting that: F y 0 M 0 s dm V d (w is positive when it acts upward) dv d w

EXAMPLE 4-4 Determine the shear and moment in the beam shown in Fig.

It is important to remember, however, that this resultant is not actual load on the beam. Shear and Moment Functions + Fy =0; 0 5 ( ) V 0 V 5.

4-7a as a function of. SOLUTION Support Reactions. The reactions at the fied support are V = 480 kn and M = 74 kn m, Fig. 4-7b.

6 m, two regions of must be considered in order to describe the shear and moment functions for the entire beam. Here is appropriate for the left.

2 EXAMPLE 4-4 Determine the shear and moment in the beam shown in Fig.4-6a as a function of Support reaction For the purpose of computing the support reactions, the distributed load is replaced by its resultant force of 0 kn/m, Fig. 4-6b. It is important to remember, however, that this resultant is not actual load on the beam. Shear and Moment Functions + Fy =0; 0 5 ( ) V 0 V Ms =0; [ ( ) ] M 0 0 Note that dm/d = V and dv/d= = w!! M EXAMPLE 4-5 Determine the shear and moment in the beam shown in Fig. 4-7a as a function of. SOLUTION Support Reactions. The reactions at the fied support are V = 480 kn and M = 74 kn m, Fig. 4-7b. Shear and Moment the Functions Since there is a discontinuity of distributed load at =.6 m, two regions of must be considered in order to describe the shear and moment functions for the entire beam. Here is appropriate for the left.6 m and can be used for the remaining segment m. Notice that V and M are shown in the positive directions, Fig. 4-7c. + Fy =0; V = 0, V = ans + MS = 0: ( ) + M = 0 M= d M / d = V d V / d = w. ().6m 6 m, Fig 4-7d. + Fy ans =0; V = 0, V = 00 ans + M = 0: ( -.8)+M = 0 M= ans These results can be partially checked by noting that when = 6 m then V = 00 kn and M = -50 kn m. Also, note that d M / d = V and d V / d = w.

3 EXAMPLE 4-6 Determine the shear and moment in the beam shown in Fig. 4-8 a as a function of. A free-body diagram of the cut section is shown in Fig. 4-8c. As above, the trapezoidal loading is replaced by rectangular and triangular distributions. Note that the intensity of the triangular load at the cut is found by proportion. The resultant force of each distributed loading and its location are indicated. Applying the equilibrium equations, we have + Fy + M =0; 75-0-[ (0)( 9 )] - V = 0 V = = 0: -75+(0)( ans )+[ (0)( 9 )] M= ans + M = 0

4 4. Shear and Moment diagram for a beam - If the variations of V and M as functions of are plotted, the graphs are termed the shear diagram and moment diagram, respectively. - Distribute loads will be considered positive when the loading acts upward. The distributed loading has been replaced by a concentrated force ( ) from the right end, where0 < <. For eample, If ()is uniform, = Apply equilibrium equation. () () at that acts at a fractional distance As noted, Eq. 4- states that the slope of the shear diagram at a point (dv/d) is equal to the intensity of the distributed load ()at the point. Likewise, Eq. 4- states that the slope of the moment diagram (dm/d) is equal to the intensity of the shear at the point. As noted, Eq. 4- states that the change in the shear between any two points on a beam equals the area under the distributed loading diagram between two points. Likewise, Eq 4-4 states that the change in the moment between two points equals the area under the shear diagram between the points. 4

4- and 4-4 cannot be used. + Fy =0 M 0 = 0 If F is, V is negative. Shear diagram jumps downward.

5 Note: Eq 4- and 4- cannot be used at points where a concentrated force acts since these equations do not account for the sudden change in shear at these points. Similarly, at points where a couple moment is applied, eq. 4- and 4-4 cannot be used. + Fy =0 M 0 = 0 If F is, V is negative. Shear diagram jumps downward. If F is, V is positive. Shear diagram jumps If M is If M is, M=. Moment diagram jumps upward., M= -. Moment diagram jumps downward must understand 5

6 Procedure for Analysis Support reaction dv shear diagram d Moment diagram dm =V d 6

Types of Structures & Loads

Structure Analysis I Chapter 4 1 Types of Structures & Loads 1Chapter Chapter 4 Internal lloading Developed in Structural Members Internal loading at a specified Point In General The loading for coplanar