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

6

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

Lecture 9-10

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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.

Lecture 9-10

9

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.

11

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 <x-a>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.

<x-a > 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 <x-a>, noting the meaning of the functions in the angular bracket that <x-a> = 0 if x ≤ a and <x−a> = (x−a) if x>a. In other words, the function <x-a> 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 – P<x-a>0 + S = 0 where the second term wo uld be zero if x ≤ a. Thus S = -RA + P<x-a>0 = -P(L-a)/L + P<x-a>0 . 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.

13

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<x-a>, 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 + P<x-a>0 ; M = P.x(L-a)/L – P<x-a>. These internal actions can be evaluated at various points. For 0<x0 . From this exercise, we see that by using discontinuity functions of the type <x-a>n we can allow for the discontinuity in loading to be incorporated into two single mathematical formulae for the shear force and bending moment. This procedure is particularly useful in the calculation of beam deflection, where integration of the bending moment expression will be required. In this example, the singularity term <x-a>0 with zero power merely acts as a switching function to include/exclude the effect of point load on the shearing force. The singularity function with a power of unity <x-a> on the other hand acts both as a switch and a multiplier (lever arm for the force, in the case of bending moment). Case 4: A simply supported beam subject to a concentrated moment For overall equilibrium, taking moments about A in the anti-clockwise direction gives: =0 RC = M0 /L

B

A

For vertical equilibrium, RA = -RC = - M0 /L

a

A negative reaction implies a downward force (a holding down force). It is worth noting here that simple supports (rockers and/or rollers) shown with triangular knife edges can exert force in either direction. i.e.

Figure 9.33

Lecture 9-10

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L-a

Figure 9.32 B

A RA

C M0

a is equivalent to

C M0

L-a

RC

For the segment shown, For vertical equilibrium, = 0 giving S = - RA = M0 /L Taking moments about the cut gives: M - M0 - RA(x) = 0 M = M0 + RA(x) = M0 - M0 (x/L) This means for, xa, M = M0 (1-(x/L))

RA

x B

A RA

C M0 RC

L-a

a

M0 /L

There is a jump in bending moment diagram at B where there is a concentrated moment loading. However the slope of the bending moment diagram on either side of B are the same. The shear force diagram does not have any change at B.

Shear Force Diagram -M0 (1-(a/L))

The singularity function that appears in the bending moment expression has a power of zero, because it only acts as a switch. This is similar to the use of <x-a>0 in Case 3 to include/exclude the effect of a point load on the shearing force.

-M0 a/L Bending Moment Diagram Figure 9.34 Bending Moment B Diagram C

A

9 kN

Σ MA= 0 gives:

1m

3m

Therefore, RB = 12 kN.

RA

RB

9 kN

RB

9 kN

M

Σ Fy = 0 gives: RA + RB - 9 = 0 S

Therefore RA = - 3 kN. For the segment shown, Σ Fy = 0 gives: -S+RB -9 = 0 gives S = RB -9 = 12

M <x-a>

a

At x = 0, M = 0 At x = a-δ (just left of B), M = At x = a+δ (just right of B), M = At x = L, M = 0.

Case 5: A simply supported beam with a load on an overhanging portion

S

M0

x

<3-x> 4-x

3 kN -9

9 kN

Taking moments about the cut gives: Shear Force Diagram - RB + 9 i.e. M = RB

=0 + 9 - 9 kNm

For 0 < x < 3 m, S =

Bending Moment Diagram Lecture 9-10

15

Figure 9.35

For 3 < x < 4 m, S = At x = 0 and at x = 4 m, M = At x = 3m, M = Notes: There is a jump in shearing force at B where there is a point force due to the support. Correspondingly the slope of the bending moment diagram also has a sudden change at this point. In this example, the bending moment at the simple support at B is not zero. While a simple support implies that the support cannot provide any moment resistance, this example shows that the bending moment in a beam at a simple support is not necessarily zero. In cases where the beam continues over a simple support or at a simple support where an external moment loading is applied the beam may have some bending moment. The use of singularity function in this problem is similar to that for Case 3. In both cases the singularity occurs at a point where there is a concentrated force (which is a load in Case 3, where as it is a reaction in the present case). In both cases, the effect of a point force on shear force is taken into account by using a switching function with a power of zero 0 , and its effect on bending moment is represented by a term with a power of unity . A

Case 6: A cantilever beam subject to uniformly distributed loading for part of its span Since this is a cantilever, the bending moment and shearing force distribution may be found by considering the equilibrium of a carefully selected segment, without having to apply overall statics. By inspection we can see that one needs to make a cut and consider the free-body that is right of the cut, to avoid having to calculate support reactions at A. If we want to obtain expressions for bending moment and shearing force that are applicable throughout the beam, we need to include all points of discontinuity in the free-body diagram. This means the only logical choice is to make a cut between A and B, and consider the free-body to the right of the cut. However, obtaining expressions for bending moment and shear force from this simple segmental free-body diagram is not easy. For example Σ Fy = 0 gives:

7 kN/m

B

C

1.2 m

1m M

S

7 kN/m

1-x

2.2 - x

M

S

7 kN/m

1-x

2.2 - x

-8.4 kN Shear Force Diagram

-S - 7(2.2-x)<x-1>0 -7(1.2)<1-x>0 = 0 S = -7(2.2-x)<x-1>0 -7(1.2)<1-x>0 This implies for x<1 m, S = -7(1.2) = - 8.4 kN, and for x> 1 m, S = -7(2.2-x) = 7x-15.4 kN. -5.04 kNm A more convenient formulation is obtained by applying two equal and opposite uniformly distributed

-13.44 kNm Bending Moment Diagram

Lecture 9-10

16

Figure 9.36

loading on the unloaded segment, and writing the equation of equilibrium. Now, Σ Fy = 0 gives: -S – 7<2.2-x> + 7<1-x> = 0 This yields: S = – 7<2.2-x> + 7<1-x> Once again, for x < 1 m, S = -7(2.2-x) + 7 (1- x) = -7 (1.2) = -8.4 kN and for x > 1 m, S = -7(2.2-x) = 7x-15.4 kN Σ Mcut = 0 gives: M+ 7<2.2-x> <2.2-x> /2 - 7<1-x><1-x> /2 i.e. M = -(7/2)(<2.2-x>2 -<1-x>2 ) For x < 1 m, M = -(7/2) ((2.2-x)2 -(1-x)2 ) = -7 (1.2)(x) - (7/2)(3.84) For x > 1m, M = - (7/2) (2.2-x)2 Evaluation: At x = 0, S = – 7<2.2-0> + 7<1-0> = -8.4 kN; M = (7/2)(<2.2-0>2 -<1-0>2 ) = -13.44 kNm. At x = 1 m, S = – 7<2.2-1> + 7<1-1> = -8.4 kN; M = (7/2)(<2.2-1>2 -<1-1>2 ) = -5.04 kNm. At x = 2.2 m, S = – 7<2.2-2.2> + 7<1-2.2> = 0; M = (7/2)(<2.2-2.2>2 -<1-2.2>2 ) = 0 The bending moment varies linearly between A 7 kN/m A and B and parabolically between B and C. B C The shear force diagram varies linearly, with a break in slope at B. 1m

Alternative Approach: The above problem could have also been solved by making a cut between B and C and considering the equilibrium of the segment that is left of the cut (see Figure 9.37). This would have necessitated calculation of reactions at A using overall equilibrium, but the use of positive and negative loading between A and B could have been avoided.

MA

RA MA

1.2 m 7 kN/m

x 7 kN/m

M S

Consider the overall equilibrium of the beam. Summing the forces vertically gives: RA – 7(1.2) = 0 i.e. RA = 8.4 kN. Taking moments about A gives: MA + 7(1.2)((1+0.6) = 0 i.e. MA = -13.44 kNm. Considering the segment from A to the cut, For vertical equilibrium, RA + S –7<x-1> = 0 S = -RA + 7 <x-1> = -8.4 + 7<x-1> For rotational equilibrium, M – MA - RA (x) + 7 <x-1><x-1>/2 = 0

Lecture 9-10

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RA

<x-1> Figure 9.37

i.e. M = MA + RA (x) - 7 <x-1>2 /2 = -13.44 + 8.4 x - 7 <x-1>2 /2 Although these expressions are different from those obtained previously using the alternative free-body, they are equivalent, and the resulting bending moment and shear force distribution are the same. A question often arises when choosing the most convenient freebody when a beam is subject to discontinuous distributed loading, and that is when should one apply equal and opposite loading. A simple answer to this is that all distributed loading should end at the cut. For example in the first approach to the present case where the cut was made between B and A, where there is no distributed loading, positive and negative loading were added to the system up to the cut. However in the alternative approach the distributed loading continues up to the point of cut, and therefore there was no need to add positive and negative loading. This point will be illustrated again using another example later on. Relationship between load, shearing force and bending moment By inspecting the bending moment and shear force diagrams for the various cases treated so far, one can deduce a relationship between these induced actions. That is, the slope of the bending moment diagram is equal to and opposite of the shearing force. This may be shown analytically, by considering the equilibrium of an infinitesimal element of a beam as follows. w(x) w(dx) S+δS M

x

δx

M+δM

S Figure 9.38

δx

Considering the equilibrium of element δx, Σ Fy = 0 gives:S + (S + δS) – w(δx) = 0 This gives: δS = w(δx) i.e δS/ δx = w Taking the limit, as δx→ 0, δS/ δx →dS/dx Therefore, dS/dx= w That is, the slope of the shearing force diagram is equal to the intensity of the applied loading. Special Case: If a prescribed force is applied over a very small length of the beam, w → ∞ Then, dS/dx → ∞ This means at a point where a concentrated (point load) is applied, the shear force diagram will have an abrupt jump. This can be seen in some of the examples considered earlier. For rotational equilibrium, summing moments in the clockwise direction about the right side cut gives: M- S (δx) - (M +δM) – w(δx) (δx)/2 = 0 Lecture 9-10

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This gives: δM = - S (δx)– w(δx) (δx)/2 Dividing both sides by δx, δM/ δx = -S - w(δx)/2 As δx → 0, w(δx)/2 is negligible compared to S. Therefore, δM/ δx → dM/dx = - S That is, the rate of change of bending moment with respect to the axial co-ordinate, which is represented by the slope of the bending moment diagram is equal in magnitude to the shearing force, but with opposite sign. This can also be observed in the bending moment and shear force diagrams for the examples that have been considered so far. This result is also useful to find the maximum bending moment in a structure, since the shear force will be zero when the bending moment is maximum or minimum. In the case of a point load, the bending moment will be maximum or minimum if the shear force abruptly changes through zero. From the above relationships between shear force, bending moment and the load intensity the following conclusions may be drawn: In shear force diagrams, There will be a sudden discontinuity (jump) in the value of the shearing force at a section where a concentrated load is applied, but the slope of the SFD does not undergo any change. There will be no sudden change in the value of the shearing force, but there will be an abrupt change in the slope of the curve at a section beyond which a distributed load is applied. There will be no change in the value of the shearing force, nor a change in the slope of the curve at a section where a concentrated moment is applied. In the case of bending moment diagrams, There will be no sudden change in the value of the bending moment, but there will be an abrupt change in the slope of the curve at a section where a concentrated load is applied. There will be no change in the value of the bending moment, nor a change in the slope of the curve at a section beyond which a distributed load is applied. There will be a sudden discontinuity (jump) in the value of the bending moment at a section where a concentrated moment is applied, but the slope of the BMD does not undergo any change. This observation is not obvious from the analysis presented earlier, but it is easy to see that including a concentrated moment in an infinitesimal segment will alter δM by the magnitude of the applied moment. Finding the shear force and bending boment using integration Integrating the differential relationship between bending moment, shear force and load gives: S2 -S1 =

∫

x2

x1

wdx which means the change in shear force between two points is equal to the

area of the loading diagram between those points. M2 -M1 = ∫ − Sdx which means the change in bending moment between two points is equal x2

x1

to the area of the shear force diagram between those points. It should be noted that these equations do not apply at points where there are point loads or moments. Where there is a point force the shear force diagram will have a jump and where there is a point moment the bmd will have a jump. Exercise: Solve the examples in this section, usi ng this method.

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Case 7: hinge

Beams connected by a

Hinge

A

4 kN/m C

If two beams are joined by a hinge B (i.e. they are free to rotate relative to each other), then the bending moment 2m 1m at the hinge will be zero, unless there is an externally applied moment, or Figure 9.39 one of the beams were to continue beyond the joint. In such cases, the structures may be considered as two (or more, if there are two or more joints) substructures. In the example shown in the figure, a cantilever beam AB is hinged to a beam BC which is simply supported at C. Both beams end at B, and are not subject to any applied moment at the hinge. Therefore the bending moment in both beams at B is zero. The hinge only provides a transverse reaction. Taking the reaction at the hinge as RB, the free-body diagrams for the two beams may be sketched. To sketch the free-body diagrams, one does not need to know the actual direction of the reaction at B. Any direction may be chosen, but care must be taken to ensure that the directions marked satisfy Newton's third law. The directions of the reactions force must be opposite as shown below. MA

4 kN/m

A

4 kN/m

B

C

B RB

RB

RA

RC 2m

1m Equal & opposite Reactions (Newton's Third Law)

(b) Simply Supported Beam (a) Cantilever For equilibrium of (b), taking moments Figure 9.40 Summing the forces transversely gives: about B gives: - RB + RC - 8 = 0 RC (2) - 4(2)(1) = 0 Therefore RB = - 4 kN. giving RC = 4 kN. Now this problem may be solved either individually, or by writing the equations of equilibrium for the following free-body:

A

For equilibrium, ↓ S - RC + 4(3-x) = 0 gives S = 4x-8 Also,

S x

C B

RC (3-x) Figure 9.41

M - RC(3-x) + 4(3-x)2 /2 = 0 gives M = RC(3- x) - 4(3-x)2 /2 =- 6+8x-2x2 At x = 0, S = - 8 kN; M = -6 kNm. At x = 1m (at the hinge), S = -4 kN; M = 0 (as expected)

Lecture 9-10

4 kN/m

M

20

At x = 3m, S = 4 kN; M = 0 (as expected) The shear force varies linearly and the bending moment varies parabolically. To find the maximum/minimum bending moment, we first need to locate the point at which it occurs. a=2m 4 kN A Shear Force Diagram

-4 kN

-8 kN

2 kNm Hinge

-6 kNm

Bending Moment Diagram Figure 9.42

Let x = a for M to be maximum/minimum. Since S = - dM/dx, S(a) = 0 i.e. 4a-8=0 a = 2 m. M (a) = -6 + 8 (a) - 2 (a)2 = -6 + 16 - 8 = 2 kNm. Note that there is no discontinuity in loading at B, and there are no singularity functions in S and M. Case 8: Simply Supported Beam Subject to Distributed Loading in a Middle segment This example illustrates how singularity functions may be used to include the effect of distributed loading acting only for some parts of a beam. For overall equilibrium, Σ M @ A gives:

6 kN/m

A

D C

B 1m

6 kN/m

RD (5) - 6(2)(2.5) = 0 RD = 4.8 kN Σ Fy = 0 gives: ↑ RA + RD - 6(2) =0 Therefore RA =7.2 kN.

RA

RD 6 kN/m

B

A

S M

RA

C

To use singularity functions, a section of the beam including all points of discontinuity should be selected. Let us make a cut between C and D and consider the free-body left of the cut. In

Lecture 9-10

2m

2m

<x-3> <x-1> x Figure 9.43

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this case, as in the first approach to Case 7, we do need to apply equal and opposite distributed loading between C and the cut. Σ Fy = 0 gives: ↑ RA + S - 6<x-1> + 6 <x-3> = 0 S = - RA + 6<x-1> - 6 <x-3> = -7.2 + 6<x-1> - 6 <x-3> Σ M @ the cut in a clock wise sense gives: RD (x) - 6<x-1><x-1>/2 + 6 <x-3> <x-3>/2 - M = 0 Therefore M = RD (x) - 3<x-1>2 + 3 <x-3>2 = 7.2 x - 3<x-1>2 + 3 <x-3>2 Evaluation: Between A and B, as x< 1, <x-1> = <x-3> =0, and S = -7.2 + 6<x-1> - 6 <x-3> = -7.2 kN M = 7.2 x - 3<x-1>2 + 3 <x-3>2 = 7.2 x. At A, M = 0 and at B, M = 7.2 kNm. Shear force is constant, and the bending moment will vary linearly. Between B and C, 1<x<3, <x-1> = (x-1) and <x-3> = 0. S = -7.2 + 6(x-1) - 6 (0) = 6x -13.2 kN. Linear variation. At x = 1m, S = -7.2 kN and at x = 3 m, S = 18-13.2 = 4.8 kN. M = 7.2 x - 3(x-1)2 + 3 (0) = -3 x2 + 13.2 x - 3. A parabolic variation. At x = 1 m, M = -3 + 13.2 -3 = 7.2 kNm. At x = 3 m, M = -3 (3)2 + 13.2 (3) - 3 = 9.6 kNm. The shear force has changed sign between B and C, and that means the maximum bending moment occurs between these points. Let x = a, for M to be a maximum. Since S (a) = 0, 6 a = 13.2 = 0. i.e. a = 2.2 m. Maximum bending moment is obtained by substituting x = 2.2 into the expression for M. This gives: M = -3 (2.2)2 + 13.2 (2.2) - 3 = 11.52 kNm.

4.8 kN Shear Force Diagram

Between C and D, 3<x<5, <x-1> = (x-1) and <x-3> = (x3). S= -7.2 + 6(x-1) - 6 (x-3) = 4.8 kN (constant). M = 7.2 x - 3(x-1)2 + 3 (x-3)2 = - 4.8 x + 24 (linear).

-7.2 kN

7.2 kNm

At x = 3 m, M = 9.6 kNm. At x = 5 m, M = 0 (as expected).

11.52 kNm 9.6 kNm

Bending Moment Diagram

Figure 9.44 Note: In evaluating the bending moment and shearing forces, it is not necessary to go through the algebraic manipulations to change the form of the equation between various sections of the beam. However, in this example this has been done to illustrate the shapes of the diagrams. All evaluations can be done using following equations: S = -7.2 + 6<x-1> - 6 <x-3> M = 7.2 x - 3<x-1>2 + 3 <x-3>2

Lecture 9-10

22

It may be noted that the shear force remains constant when the intensity of loading is zero, and the bending moment has a constant slope. In segments subject to a uniformly distributed loading, the shear force varies linearly, and the bending moment diagram is parabolic. This problem can also be solved by considering a freebody right of a cut made between A and B. Although the expressions for bending moment and shearing force may be of different forms, their values will be the same. Case 9: Symmetrical and Anti-symmetrical Loading on Symmetrical Structure

10 kN

10 kN

6 kN/m B

A

1m

C

D

1m

2m

10 kN

10 kN

RB

RC

When a symmetrical structure is subject to symmetrical load 10 kN M S the resulting bending moment diagram will be symmetrical and the shear force diagram RC RB will be anti-symmetrical. If on <x-3> the other hand the loading is <x-1> anti-symmetrical, the bending moment diagram will be anti- 10 kN x symmetrical and the shear force diagram will be symmetrical. For overall equilibrium, -10 kN Σ M @ B gives: RC (2) - 10(3) Shear Force Diagram + 10(1) = 0 RC = 10 kN Σ Fy = 0 gives: RB + RC - 20 =0 Therefore RB = 10 kN. These results could have been -10 kNm obtained by applying Bending Moment Diagram symmetry. i.e. Since RB = RC and Figure 9.45 RB + RC - 20 =0, RB = RC = 10 kN. Σ Fy = 0 gives: RB<x-1>0 +RC<x-3>0 +S-10 = 0 S = -RB<x-1>0 -RC<x-3>0 +10 Σ M @ the cut gives: M-RC<x-3>-RB<x-1>+10(x)=0 M=RC<x-3>+RB<x-1>-10(x) Exercise: Change the direction of one of the 10 kN loads, which would make the loading anti-symmetrical (symmetrical but on one side of centre- line it is opposite in sign). This would result in symmetrical shear force diagram and anti-symmetrical bending moment diagram.

Lecture 9-10

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