Transverse Loading Calculation of Bending Moment and Shear Force in Beams

Transverse Loading: Calculation of Bending Moment and Shear Force in Beams Skeletal members that primarily carry loading applied perpendicular to their axes (transversely or laterally) will be referred to as ‘beams’. As they carry transverse loading they undergo bending or flexural deformations and to some extent shearing. If the crosssectional dimensions of a beam are small compared to their span (length of the beam undergoing bending) it may be classified as a slender beam. In most practical situations, the beams may be treated as slender. Slender beams undergo flexural deformations where one side of the beam gets stretched while the other side gets shortened as shown in Figure 9.13.

shortening

elongation

Flexural deformation

Shearing & Flexure

Figure 9.13. There will also be a shearing type deformation, which is considerable in beams that are short (cross-sectional dimensions being comparable to the span). The deformations shown in the diagrams above are grossly exaggerated, and in practice the displacements are very small. Beams are used in most structures and machines, and the effect of transverse loading on beams is often the main criteria in structural design. Typical examples include bridges, roof structural elements (purlins, rafters), floor-beams, machine elements such as levers, cranks, manipulator arms etc. Even members that predominantly carry torsional loading such as transmission shafts are susceptible to flexural deformations, and knowledge of their flexural behaviour is needed in calculating the whirling speeds (the speed at which the shaft has a tendency to whirl) of shafts. Therefore, it is important to be able to calculate the internal actions in beams due to transverse loading. Types of Induced Actions Let us consider a beam that is clamped at one end and unsupported at the other end as shown in Figure 9.11. This is called a cantilever beam. Let the beam be subject to an upward force P (applied load) at distance ‘a’ from the clamp. A sketch of its deflected shape (exaggerated) is also shown. C A

B

C

B A

P

a

Deflected Shape (exaggerated)

Cantilever Beam Figure 9.14.

Lecture 9-10

P

a

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The reactions at A may be calculated by considering the overall equilibrium. First let us sketch the overall free-body diagram. RA The free-body diagram in Figure 9.15 shows the applied MA force P and two induced reactions at A, the transverse A C B force RA and the reaction moment MA. The actual sense P of the reactions will be known only after solving a equations of equilibrium. The directions shown were chosen arbitrarily. Summing the forces in transverse Figure 9.15. direction gives: Overall Free-body diagram ↑ RA + P =0 RA = -P. This means that the actual sense of the reaction RA is opposite to the one shown in the freebody diagram. Similarly summing the moments about A gives: MA - P.a = 0 or -MA + P.a = 0 which yields MA = P.a. To find the internal actions in the beam, we need to apply the method of sections. Let us make a transverse cut in the beam at distance x from A, so the cut lies between A and B and consider one of the free-bodies shown in Figure 9.16:

MA A

C

B

(a-x)

x

P

Figure 9.16 Figurediagrams 1.2.18 are incomplete. The actions from one free-body onto Clearly the above free-body the other have to be inserted to complete these free-body diagrams. It can be seen that a transverse force and a moment will have to be inserted at the cut, to maintain the equilibrium of these free-bodies. The force induced acts in a direction (transverse) that is parallel to the cut and therefore it is referred to as a shearing force. The moment is associated with the bending of the beam and is therefore called a bending moment. Sign Convention for Shear Force and Bending Moment At this stage we need to choose sign conventions for the shear force and the bending moment. If we show the shear force as acting in the downward direction on one of the freebodies, by Newton’s third law, the shear force on the other free-body must be shown as acting in the upward direction. One possible convention is to show the shear force on the right- hand side of the left segment as acting in the upward direction as shown in Figure 9.17. To be consistent with Newton’s third law, we should then show the shear force acting on the left- hand side of the right-hand side free-body in the downward direction. Similarly, if we Lecture 9-10

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show the moment on the right-hand side free-body in a clock-wise sense, the bending moment on the left-hand-side free-body should be shown in the anti-clockwise sense. Therefore Figure 9.17 shows one possible consistent set of sign conventions for the induced actions. The shear force is denoted by SAB and the bending moment is denoted by MAB. As in the case of axial force, the subscript ‘AB’ refers to the segment in which the internal actions act. Applying Equilibrium Equations : RA

SAB MAB

MAB

MA A

C

B SAB (a-x)

x

P

Figure 9.17 Completed free-body diagrams of two beam segments Now we can apply the equations of equilibrium to determine the unknown actions. First let us find the shear force. Summing the transverse forces acting on the right- hand side freebody we get: ↑ - SAB + P = 0. Therefore SAB = P. The same result could have been obtained by summing the forces in the downward direction, or by considering the equilibrium of the other free-body as shown below: Summing the forces in the downward direction gives ↓SAB - P = 0 which also gives the same result for the shear force. Alternatively, taking the left- hand side free-body gives ↑ SAB + RA =0 giving SAB = -RA; But we already have RA = -P. Therefore SAB = -(-P) = P. Once again we get the same result. To find the bending moment, we can sum the moments for one of the free-bodies. Let us consider the right-hand side free-body and take moments in the clockwise direction about a point through the cut. We get, MAB – P(a-x) = 0 giving MAB = P(a-x). Thus we have an expression for the bending moment. We could have taken moments about any other point, for example about B or C. There is however an advantage in taking moments about a point on the cut- face. As the induced shear force passes through the cut, if moments were taken about any point that does not lie on the cut- face, the contribution from the moment due to the shear force also enters the equation. Since the shear force is also an unknown, any error in the calculation of the shear force would also introduce an error in the calculation of the unknown bending moment. Having obtained the expressions for the shear force and bending moment in segment AB, we can now move to segment BC. If we make a cut between B and C and consider the freebody right of the cut, there will be no applied force in that segment. Therefore, the bending moment and shear force in BC must be zero

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The results obtained may be sketched as shear force and bending moment diagrams.

P.a

P

A

B

C

A

B

Shear Force Diagram

C

Bending Moment Diagram

Figure 9.18 Shear Force Diagram compression

In interpreting the shear force and bending moment diagrams, one needs to refer to the sign convention used. The convention used here may be shown in a single diagram as follows:

Shear Force acts in the upward direction (positive y-direction) on the right- hand side face (positive face) and downward on the left- hand side face (negative face) of a beam segment. The pair Figure 9.19 of shear forces would form a couple in the anti-clockwise sense, and for this reason this convention for shear force may be referred to as ‘anti-clockwise shear positive’ convention. Tension

Bending Moment acts in the clockwise direction on the left-hand side face and anti-clockwise on the right- hand side face. This bending moment causes compression at the top and tension at the Figure 9.20 bottom corresponding to a sagging type curva ture. This shape may be associated with the mouth of a happy face, and those using this sign convention may want to remember that a happy face type bending is positive. There are four consistent sets of sign convention for shear force and bending moment. They are shown in Figure 9.21.

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2. Sagging moment +ve Clockwise shear +ve

1. Sagging moment +ve Anti-clockwise shear +ve

4. Hogging moment +ve Clockwise shear +ve

3. Hogging moment +ve Anti-Clockwise shear +ve Figure 9.21

Throughout this text, the first set of sign conventions will be used. However, some of the interactive multi- media tutorial modules are designed to work in any of the above conventions. The users may select any one of the four conventions. The following examples illustrate the application of method of sections for various loading and support conditions.

6 kN/m A

2m

B

Case 1: A cantilever subject to a uniformly distributed load MAB 6 kN/m (UDL) A 2m long cantilever beam is subject to a uniformly distributed load of 6 kN/m as shown in Figure 9.22. The bending moment SAB (2-x) m and shear force distributions in the beam are required. Let us make an imaginary cut at distance x from the clamp, and consider the equilibrium of one of the free-bodies. In this case, it is (2-x)/2 m convenient to choose the free-body that is right of the cut because there will be only two unknown actions on it. In the free-body left 6(2-x) kN MAB of the cut, the reactions at the clamp (a force and a moment) will be required. Although they can be obtained from overall equilibrium equations, if we only need the shear force and bending (2-x) m moment, it is simpler to apply equations of equilibrium to the SAB right-side free-body which is shown here. The length of the freeFigure 9.22 body is (2-x) m. The induced actions MAB and SAB are shown according to the sign convention chosen (Set 1). Since the loading on the segment is uniformly distributed, it is equivalent to a net force of 6(2- x) kN acting at a distance of (2-x)/2 m from the cut, as shown in the equivalent free-body diagram. Summing the forces in the transverse direction gives: ↑ Therefore SAB =

= 0. kN

Taking moments about the cut gives: =0 Lecture 9-10

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Therefore MAB = At x = 0, SAB = -6(2-0) = -12 kN; MAB =-6(2-0)2 /2 = -12 kNm. At x = 2m, SAB = -6(2-2) = 0; MAB =-6(2-2)2 /2 = 0. The shear force varies linearly with x while the bending moment varies parabolically as shown below:

-12 KN

-12 KNm Shear Force Diagram

Figure 9.23

Bending Moment Diagram

Case 2: A simply supported beam subject to a uniformly distributed load (UDL) A simply supported beam of length L is subject to a distributed load of intensity w per unit length. The bending moment and shear force distributions are required. In applying the method of sections, whichever side of the cut we choose to consider, the reaction from one of the supports will act on the free-body. Therefore we need to determine one of the reactions first, using the overall equilibrium equation. w w The only reactions B at the supports are A A B lateral forces, as RA RB there are no horizontal forces L and the supports Figure 9.24 cannot sustain moments. By taking moments about one of the support points, the reaction at the other support may be found. However, in this case, it is simpler to take half the net applied force as the reaction at either end, since the structure is symmetrical. The fact that there is a roller support at one end and a rocker at the other end does not alter lateral symmetry, as the difference between the horizontal support conditions will only affect horizontal forces. w.x SAB Using symmetry, RA = RB = wL/2. For a segment of the beam having length x, For transverse equilibrium,

MAB RA x

↑ giving SAB =

= 0, Figure 9.25

= w(x-(L/2))

Taking moments about the cut, = 0, which gives MAB =

Lecture 9-10

= wx(L-x)/2.

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The shear force at x = 0 is –wL/2 and at x = L it is + wL/2. The bending moment is zero at x=0 and at x=L, and at x=L/2, M = wL2 /8. From the above equations we can sketch the shearing force and bending moment diagrams. In this example, the structure and the loading are symmetrical, and the bending moment wL2 /8

wL/2

-wL/2 Bending Moment Diagram

Shear Force Diagram Figure 9.26

diagram is also symmetrical (the shape on one side of the structure is a mirror image of the shape on the other side). The shear force diagram is anti-symmetrical, that is symmetrical in shape but with opposite signs. Use of singularity functions for beams subject to discontinuo us loading In cases where the loading is not continuous over the length of the beam, the shear force and bending moment in the various segments will have different expressions. To obtain these, one can make the cut in the various segments, and obtain the equations of equilibrium for each of the free-bodies. However, by using singularity functions such as n it is possible to allow for the discontinuities and obtain expressions for the induced actions that are applicable throughout the beam. Case 3: A simply supported beam subject to a concentrated (point) load a a Let us now consider P a simply supported P beam of length L, C A A subject to a point C RA RC load P applied at a distance ‘a’ from the L left support as L shown in the Simply supported beam Overall free-body diagram diagram. For overall equilibrium, under a point load taking moments Figure 9.27 about the left support in anti-clockwise sense gives: RC L – P. a = 0 which yields RC = P a/L Similarly by taking moments about C it can be shown that RA = P (L-a)/L Alternatively, substituting RC = P a/L into the lateral equation of equilibrium RA+RC-P=0 also gives the same expression for RA. To find the induced actions in the beam let us apply the method of sections by making a cut at distance x from A. First let us consider the case where x < a, so that the applied load P would not appear on the left-side free-body. Lecture 9-10

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↑Σ F = 0 gives RA + SAB = 0 ⇒ SAB = - RA = - P (L-a)/L. Taking moments about the cut in the clockwise direction,

SAB RA

RA.x – MAB = 0 which yields, MAB = RA.x = P.x(L-a)/L.

MAB

x

Figure 9.28

At x = 0, MAB = 0 and just left of B, MAB = P.a(L-a)/L The above derivations are applicable for x < a, and for x>a we need to sketch another free-body diagram, this time including the applied load too.

a

P SBC MBC

RA

For this freebody, transverse equilibrium equation is: RA – P + SBC = 0 i.e. SBC = -RA + P =-P(L-a)/L + P = P.a/L

x Figure 9.29

For rotational equilibrium, by taking moments about the cut, we have: = 0 giving MBC = = Pa(L-x)/L Note that the subscripts for the shear force and the bending moment have now been changed to BC. It is possible to write down expressions for the shear force and bending moment that would be applicable for both segments.

a

S RA

Let us sketch a new free-body diagram, and this time label the distance between the cut and the load P in the last free-body as , noting the meaning of the functions in the angular bracket that = 0 if x ≤ a and = (x−a) if x>a. In other words, the function would be active if and only if the term within the bracket is positive. Let us also remove the subscripts of the shear force S and the bending moment M, with the object of obtaining expressions for S and M that are valid for the full length of the beam. Using the modified free-body diagram, we may write the transverse equilibrium equation as: RA – P0 + S = 0 where the second term wo uld be zero if x ≤ a. Thus S = -RA + P0 = -P(L-a)/L + P0 . By comparing this with the expression for SAB and SBC it may be seen that this expression for the shear force is valid for 0 ≤ x ≤ L.

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M

x Figure 9.30 P.a/L

-P (L-a) /L Shear Force Diagram P.a(L-a) /L

Bending Moment Diagram Figure 9.31

Similarly an expression for the bending moment would be obtained as:

Lecture 9-10

P

M = P.x(L-a)/L – P, once again the significance of the angle bracket being the second term in the equation would vanish if x ≤ a. This would agree with the expressions for MAB and MBC obtained previously. Summarising the results for this case, the expressions for the bending shear force and bending moment are: S = -P(L-a)/L + P0 ; M = P.x(L-a)/L – P. These internal actions can be evaluated at various points. For 0