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Slide 2 Lecture 4a Shear Force and Bending Moment Diagrams (part 2) ENR 202 Mechanics of Materials Welcome to Lecture Summary 4A. In this lecture summary, we will continue learning about shear force diagrams and bending moment diagrams. We have recently covered the sign conventions for shear force and bending moment, the definition of shear force and bending moment, and the concepts of shear force diagrams and bending moment diagrams, using the equilibrium method. We will continue with the equilibrium method now. Note that throughout all the lecture summaries for Mechanics of Materials, you will see live links, denoted by the letters W, P and V. These links point to web pages, presentations and videos which will enhance your understanding of the content. You can pause the presentation at any time to access these links, and then go back to the presentation when you have finished looking at them.
Slide 3 Requirement for students ENR202 4a -- Slide No. 3 To be able to draw SFD and BMD quickly and correctly The aim of this presentation is to teach you to draw shear force diagrams and bending moment diagrams correctly.
Slide 4 General procedure to calculate shear force and bending moment in beams Calculate reaction forces Draw free body diagram with cutting at the calculating point. For cutting at continuous points, there are three internal actions: normal force(usually =0), shear force and bending moment. Calculate shear force and bending moment at important points based on equilibrium equations ENR202 4a -- Slide No. 4 First, we will go through the procedure for analysis again. The first step is to calculate the reaction forces at the supports. This step is not difficult, but it is very important, because if you calculate the reaction forces incorrectly, your shear force and bending moment values will also be definitely wrong. After you have calculated the reaction forces, you can cut at any cross section of the beam, and draw the free body diagram of one of the beam segments. As you know, we have internal actions at the cutting cross section. Once you draw the free body diagram, then you can calculate the internal actions. You can calculate the shear force functions and the bending moment functions by using equilibrium equations.
Slide 5 Exercise 1: point load on simply supported beam ENR202 4a -- Slide No. 5 Draw SFD and BMD for beam ABC In past lectures, we have drawn Shear Force Diagrams and Bending Moment Diagrams for a simply supported beam subjected to a uniform distributed load. Let s look at this example of that simply supported beam, subjected to a point load of 30 kilo Newtons acting 2 meters from left end of the beam at Point B, as shown in the figure. The total length of the beam is 6 meters. Pause this presentation and try to draw the shear force diagrams and the bending moment diagrams for this simply supported beam. The solution is on the following four slides.
Slide 6 Exercise 1: Solution (1) ENR202 4a -- Slide No. 6 Step 1: Reaction forces M C =0 R A *6m-30kN*4m=0 R A =20kN M A =0 30kN*2m-R C *6m-=0 R C =10kN Check F y =0(???) R A +R C -30kN=0 (!) OK! The first step is to calculate the reaction forces at the supports. You can calculate reaction forces at A and C. We have two reaction forces, RA and RC, so we need two equilibrium equations. The first equilibrium equation is that the sum of all force moments about C is equal to zero. We have two force moments about C. The first is the moment of reaction force RA, which is RA times 6 meters in a clock wise direction. The second is the moment of point load 30 kilo Newtons about C, which is 30 kilo Newtons times 4 meters in an anti-clockwise direction. So the reaction force RA is 20 kilo Newtons. The second equilibrium equation is that the sum of all force moments about A is equal to zero. We have two force moments about A. The first is the moment of reaction force RC about A, which is RC times 6 meters in an anti-clockwise direction. The second is the moment of point load 30 kilo Newtons about A, which is 30 kilo Newtons times 2 meters in a clockwise direction. So we can calculate that the reaction force RC is 10 kilo Newtons. To check the accuracy of this result, consider one more equation: that is, that all forces in a vertical direction are equal to zero. The sum of reaction forces RA and RC is equal to 30 kilo Newtons. So now we know that our reaction force calculations are correct.
Slide 7 Exercise 1: Solution (2) ENR202 4a -- Slide No. 7 Step 2: Free body diagram (cut at B left, x<2m) F y =0 R A -V (x) =0 V(x)=R A =20kN (1) M X =0 R A *x-m(x)=0 M(x)=R A *x=20x (2) M(x=2m)=20*2=40kN-m The next step is to cut the beam between point A and point B at a distance x from the left support. We have two internal force actions at the cutting surface. One is a shear force and the other is a bending moment. Draw the free body diagram of the left side of the beam as shown in the figure. As you know, in a free body diagram, you remove supports and replace reaction forces. The diagram will show loads on the structure and internal actions of shear force v(x) and bending moment M(x), as shown in the figure. V(x) and M(x) are positive shear and positive bending moments for the left side of the beam, as shown in the figure. This shear force v(x) is acting downward and rotates the beam segment in a clockwise direction. Due to bending moment M(x), the top part of the beam is in compression and the bottom part of the beam is in tension. You can write the equilibrium equations for the left part of the beam. The first equilibrium equation is that all forces in a vertical direction are equal to zero. We have two forces. One is reaction force RA, which is 20 kilo Newtons acting in an upward direction, and the other is an internal shear force at the cut cross section v(x), which acts in a downward direction. So, for the first equation, shear force V(x) is equal to 20 kilo Newtons, as shown in the equation 1. The second equilibrium equation is that the sum of all force moments about Point x must be equal to zero. We have three moments about Point x. The first is the moment of reaction force RA, which is 20 kilo Newtons about x, which is 20 kilo Newtons times distance x in a clockwise direction. The second is the moment of M(x), which is anti-clockwise. The third is the moment of v(x) about x, which is equal to v(x) times zero distance. Finally, we get the moment function M(x) as shown in equation 2, in terms x that means linear equation.
If you substitute x equal to 2 meters in moment equation 2, you will see that the moment at Point C is 40 kilo Newton meters.
Slide 8 Example 1: Solution (3) ENR202 4a -- Slide No. 8 Step 3: Free body diagram (cut at B right with x>2m) F y =0 R A -30kN-V(x)=0 V(x)=-30+R A =-10kN (3) M X =0 R A *x-30*(x-2)-m(x)=0 M(x=2)=40kN-m M(x)=60-10x (4) The next step is to cut the beam between points B and C at a distance x from the left support. We have two internal force actions at that cutting surface. One is a shear force and the other one is a bending moment. Now draw the free body diagram of the left side of the beam, as shown in the figure. V(x) and M(x) are positive shear and positive bending moment for the left side of beam, as shown in the Figure. This shear force v(x) is acting downward and rotates the beam segment in a clockwise direction. Due to the bending moment M(x), the top part of beam is in compression and the bottom part of the beam is in tension. You can write the equilibrium equations for this beam segment. The first equilibrium equation is that all forces in a vertical direction are equal to zero. We have 3 forces. One is reaction force RA, which is 20 kilo Newtons acting in an upward direction. The second is a point load of 30 kilo Newtons acting in a downward direction. The third is an internal shear force at the cutting cross section, which is v(x) acting in a downward direction. Finally we work out the equation for shear force V(x) which is equal to minus 10 kilo Newtons, as shown in the equation 3. The second equilibrium equation is that the sum of all force moments about Point x is equal to zero. We have four moments about x. The first is the moment of reaction force RA, which is 20 kilo Newtons about x, which is 20 kilo Newtons times distance x in a clockwise direction. The second moment of point load is 30 kilo Newtons times the distance between B and x, which is (x-2) distance in an anticlockwise direction. The third is the moment of M(x) in an anti-clockwise direction. The fourth is the moment of v(x) about x, which is equal to v(x) times zero distance. Finally, we get the moment function M(x) as shown in the equation 4; that
is, 60 minus 10 times x that means in terms x. So equation 4 is the linear equation. If you substitute x equal to 2 meters in moment equation 2, you will see that the moment at Point C is 40 kilo Newton meters.
Slide 9 Example 1: Solution (4) ENR202 4a -- Slide No. 9 SFD BMD Step 4: Draw SFD and BMD Equations 1 and 2 are valid in between point A and point B. Equations 3 and 4 are valid in between point B and point C. Therefore, the shear force function v(x) seen in Equation 1 is a constant value of 20 kilo Newtons between points A and B. So it is horizontal line in the Shear Force Diagram. The shear force function v(x) seen in equation 3 is a constant value. It is minus 10 kilo Newtons between point B and C. So it is a horizontal line with a value of minus 10 kilo Newtons in SFD. This is the shear force diagram based on the shear force function. We have zero shear force at Point B. In the bending moment diagram between point A and point B, the bending moment function of equation 2 is 20 times x. We have a linear equation, which means the bending moment diagram is an inclined straight line as shown in the figure between point A and point B. If you substitute x equal to zero into equation 1, the bending moment function is zero. If you substitute x equal to 2 meters into equation 2, the bending moment is 40 kilo Newton meters. In the bending moment diagram between points B and C, the bending moment function of equation 4 is 60 minus 10 times x. We have a linear equation which means the bending moment diagram is an inclined straight line as shown in the figure between point B and point C. If you substitute x equal to 2 into equation 4, the bending moment is 40 kilo Newton meters. If you substitute x equal to 6 meters into equation 4, the bending moment function is zero. Finally, the maximum shear force occurs between A and B as shown in the Shear Force Diagram. The maximum bending moment happens in the middle of the beam as shown in the Bending Moment Diagram.
As you know, anything drawn in a Shear Force Diagram above the x-axis is positive, and anything drawn in a Bending Moment Diagram below the x-axis is positive.
Slide 10 ENR202 4a -- Slide No. 10 W ritin g S h e a r F o rc e a n d B e n d in g Mo m e n t fu n c tio n s is q u ite a ha rd ta s k..s o is it po s s ib le to dra w S h e a r F o rc e D ia g ra m s a n d B e n d in g Mo m e n t D ia g ra m s ba s e d o n im p o rta n t po in ts w ith o u t w ritin g S h e a r F o rc e o r B e n d in g Mo m e n t fu n c tio n s...? In previous examples, you have developed shear force functions and bending moment functions so that you could draw Shear Force Diagrams and Bending Moment Diagrams. However, it is possible to calculate shear forces and bending moments at important points without writing shear force or bending moment functions. Let us try to calculate shear force and bending moment functions at important points.
Slide 11 Properties of SFDs and BMDs ENR202 4a -- Slide No. 11 Load changing point (e.g. Concentrate load) will cause the function change of SF and BM Zero UDL will cause SFD horizontal line, BMD a sloped straight line UDL will cause SFD to be a sloped straight line, BMD to be a parabola Remember these points. Concentrated load will cause the function change of shear force and bending moment. For a beam segment without any distributed loads within the span, the shear force diagram will be a horizontal line, and the Bending Moment Diagram will be an inclined straight line, as shown in Slide 9 of this lecture. The UDL will create an inclined straight line in a Shear Force Diagram and a parabolic variation in a Bending Moment Diagram, as shown in slide 31 of Lecture 3b. If there is a triangle load (an inclined straight line) acting on the beam, the shear force will be a parabolic variation. Remember that the shear force diagram increases one order extra than the load. For example, if the UDL is zero order, then the shear force is an inclined straight line, which means linear variation, which means first order. If a triangle load is acting on the beam, this is a first order load, and the shear force is a parabolic variation, which is a second order load. If there is only one concentrated load acting on the beam, the bending moment diagram will be an inclined straight line. If a UDL is acting on the beam, the bending moment diagram will be a parabolic variation. If there is a triangle load (inclined straight line) acting on the beam, the bending moment will be a cubic parabolic variation. Note that the bending moment diagram is two orders higher than the load. For example, if the UDL is zero order, the bending moment is parabolic, which is a second order variation. If there is a triangle load acting on the beam, that load is a first order load, so that bending moment is a cubic parabolic variation, which is a third order.
Slide 12 Procedure to draw SFD and BMD based on important points Calculate reaction forces Separate beam into small segments based on important points: Supporting points, loading changing points (start and end of distributed loads, concentrate loading points and concentrate moment points), zero shear force point (for BMD) Calculate SF and BM at the important points. Pay attention to concentrate loading point (SF changes at the point) ENR202 4a -- Slide No. 12 Connect V values at important points using straight lines (For a beam segment with q=0, or a constant q). Connect M values at important points with straight line (for a beam segment with q=0), or parabolic line (for a beam segment with nonzero q). This is the procedure to draw shear force diagrams and bending moment diagrams based on important points. Firstly, calculate the reaction forces at the supports, as we explained in lecture 3a. Secondly, separate the beam into small beam segments based on important points, such as the supporting point and the loading changing point. For example, this could be the start or end points for a distributed load, or the concentrated load point, or the concentrated moment point, or the zero shear force point. Use these to draw the bending moment diagram. Thirdly, you need to calculate the shear force and the bending moment at the important points. However, make sure that you consider the concentrated load point when calculating the shear force, because the Shear Force changes at the left side of the point load and at the right side of the point load. Also consider the concentrated moment point when calculating the bending moment, because the bending moment changes at the left side of the concentrated moment and at the right side of the concentrated moment. Connect all the shear force values at important points with zero UDL or UDL in the beam segment using straight lines. Connect all bending moment values at important points in the beam segment with zero UDL using straight lines and parabolic variation for non-zero UDL.
Slide 13 Exercise 2: ENR202 4a -- Slide No. 13 Step 1: Reaction forces R A =20kN R C =10kN Step 2: important points (based on FBD analysis) A: V=20kN M=0 B left: V=20kN M=40kN-m B right: V=-10kN M=40kN-m C: V=-10kN M=0 This is the same type of problem as was explained before. A simply supported beam is subjected to a point load of 30 kilo Newtons acting 2 meters from left end; that is, Point B, as shown in the figure. The total length of the beam is 6 meters. Pause this presentation and draw the Shear Force Diagram and Bending Moment Diagram for this simply supported beam. The solution is on the next three slides.
Slide 14 Exercise 2: Solution (1) ENR202 4a -- Slide No. 14 Step 1: Reaction forces R A =20kN R C =10kN Step 2: important points (based on FBD analysis) A: V=20kN M=0 B left: V=20kN M=40kN-m B right: V=-10kN M=40kN-m C: V=-10kN M=0 As you know, the first step is to calculate the reaction forces at supports; that is, at Point A and C. We have two reaction forces: RA and RC, so we need two equilibrium equations. The first equilibrium equation is that the sum of all force moments about C is equal to zero. We have two force moments about C. The first is the moment of reaction force RA about C, which is RA times 6 meters in a clockwise direction. The second is the moment of point load 30 kilo Newtons about C, which is 30 kilo Newtons times 4 meters in an anti-clockwise direction. So we can calculate the reaction force RA, which is 20 kilo Newtons. The second equilibrium equation is that the sum of all force moments about A is equal to zero. We have two force moments about A. The first is the moment of reaction force RC about A, which is RC times 6 meters in an anti-clockwise direction. The second is the moment of point load 30 kilo Newtons about A, which is 30 kilo Newtons times 2 meters in a clockwise direction. So we can calculate reaction force RC, which is 10 kilo Newtons.
Slide 15 Exercise 2: Solution (2) ENR202 4a -- Slide No. 15 Step 1: Reaction forces R A =20kN R C =10kN Step 2: important points (based on FBD analysis) A: V=20kN M=0 B left: V=20kN M=40kN-m B right: V=-10kN M=40kN-m C: V=-10kN M=0 We have three important points here. There are two support points at Point A and C, and a concentrated load at Point B. As we have discussed, if we have a concentrated load, you must consider the left side and the right side of point B to calculate the shear force. The shear force at Point A is nothing, but the reaction force at RA is 20 Kilo Newtons. We know already that there is no bending moment at the hinged support. Next we move on to point B. We have looked at how to calculate shear force and bending moment in lecture 3b, on slide numbers 14 and 15, but we will go through it again. If you want to calculate the shear force and bending moment at the left side of B, you just cut the beam at the left side of B and consider the left part of the simply supported beam segment. Draw the free body diagram. In the free body diagram, show the reaction force at support A, which is RA. The left shear force at B is left VB, coming down at the cutting point, and is a positive shear force. The moment at B at the cutting point is MB, in an anti-clockwise direction, which is positive moment. Next, apply equilibrium equations to the free body diagram. One is that all forces in a vertical direction equal zero. Finally, we can work out the left shear force at B which is equal to 20 kilo Newtons. The next equilibrium equation is that the sum of all force moments at Point B is equal to zero. Finally, we can work out that the moment at B is equal to 40 Kilo Newton meters. If you want to calculate shear force and bending moment at right side of Point B point, just cut the beam at the right side of point B and consider the left part of the simply supported beam segment. Draw the free body diagram. In the free body diagram, show the reaction force at support A as RA, a point load of 30 kilo Newtons acting down. The right shear force at B is right VB, which is coming down at the
cutting point and is a positive shear force. The moment at B at the cutting point is MB, in an anti-clockwise direction, and is positive. Next, apply equilibrium equations on free body diagram. One is that all forces in vertical direction are equal to zero. So we can work out that the right shear force at B is equal to minus 10 kilo Newtons. The next equilibrium equation is that the sum of all force moments at B is equal to zero. Finally, we can calculate the moment at B is equal to 40 Kilo Newton meters. The last important point is C. The shear force at C is nothing, but the reaction force at C is minus 10 kilo Newtons. We already know there is no bending moment at the roller support. You can try yourself to cut the beam at the left side of Point C and find the shear force and bending moment.
Slide 16 Exercise 2: Solution (3) ENR202 4a -- Slide No. 16 Step 3: Draw SFD and BMD Now show the shear force values in a Shear Force Diagram and the bending moment values in a Bending Moment Diagram. Next connect all shear force values at important points using straight lines, to get the Shear Force Diagram as in this slide. Then connect all bending moment values at important points using straight lines, and you will get above bending moment diagram.
Slide 17 Exercise 3 ENR202 4a -- Slide No. 17 A WL/2 w L B x WL/2 Draw SFD and BMD Now pause the presentation and try to complete this exercise. A simply supported beam is subjected to uniform distributed load (UDL) w throughout the length of the beam as shown in the figure. Draw the Shear Force Diagram and Bending Moment Diagram for this simply supported beam based on the important points principles. The solution is on the next two slides.
Slide 18 Exercise 3: solution (1) ENR202 4a -- Slide No. 18 A WL/2 w L B x WL/2 Step 1: Reaction forces R A =R C =WL/2 (1) Step 2: important points A: V=WL/2 M=0 B : V=-WL/2 M=0 Step 3: Draw SFD (straight line connecting A and B ) Draw BMD (Parabola between A and B ) Firstly, calculate the reaction forces at the supports. You can calculate reaction forces at A and B. This is a symmetric beam which is subjected to a symmetric loading. The resultant of the UDL is equal to w times the length of the beam. Therefore, the reaction forces at A and B are equal, and the value is Uniformly Distributed Load w times the length of the beam L divided by 2, as shown in Equation 1. Now work out how many important points there are. You should work out that there are two important points. The first is Point A, because it is a support point and also the beginning of the Uniformly Distributed Load, and other is Point B, because it is another support, and the end of the Uniformly Distributed Load. You can cut the beam at the right side of A to find the shear force, which is nothing. However, the reaction force at A is equal to w times L divided by 2. You can cut the beam at the left side of Point B to find the shear force, which is nothing. However, the reaction force at B is equal to minus w times L divided by 2. We already know that there is no bending moment at Point A and B because of the hinged and roller supports respectively.
Slide 19 ENR202 4a -- Slide No. 19 Exercise 3: solution (2) w A WL/2 L B x WL/2 SFD WL/2 + - -WL/2 BMD + 2 wl 8 + Considering one more important point with 0 shear force, M=wL 2 /8 Now show the shear force values in the Shear Force Diagram and bending moment values in the Bending Moment Diagram. Next, connect all shear force values at important points using straight lines. From this, you will get the Shear Force Diagram that you see above. As already explained in slide number 13 of this lecture, the zero shear force point is one more important point to draw a bending moment diagram. The zero shear force occurs at the middle of the beam, so you need to calculate the bending moment at the middle of the beam. Cut at the middle of the beam and draw the free body diagram and calculate the bending moment at the cutting cross section. Next, connect all bending moment values at important points by a parabola because the load is a Uniformly Distributed Load. You should get the bending moment diagram that you can see above. Note that in this exercise, we haven t found the shear force function and bending moment function to draw shear force diagram and bending moment diagram. We have just calculated shear force and bending moment values at important points only.
Slide 20 ENR202 4a -- Slide No. 20 THANK YOU This is the end of lecture 4A. Thank you for your attention.