Shear and moment diagram - wikipedia Shear and moment diagram Shear and Bending moment diagram for a simply supported beam with a concentrated load at mid-span. ( right ) Shear and bending moment diagrams are analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of a structural element such as a beam. These diagrams can be used to easily determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure. Another application of shear and moment diagrams is that the deflection of a beam can be easily determined using either the moment area method or the conjugate beam method. Contents 1 Convention 1.1 Normal convention 1.2 Alternative drawing convention 2 Calculating shear force and bending moment 2.1 Step 1 : Compute the reaction forces and moments 2.2 Step 2 : Break beam into segments 2.3 Step 3 : Compute shear forces and moments - first piece 2.4 Step 4 : Compute shear forces and moments - second piece 2.5 Step 5 : Compute shear forces and moments - third piece 2.6 Step 6 : Compute shear forces and moments - fourth piece 2.7 Step 7 : Compute deflections of the four segments 2.8 Step 8 : Apply boundary conditions 2.9 Step 9 : Solve for M and R 2.10 Step 10 : Plot bending moment and shear force diagrams 2.11 Relationship between shear force and bending moment 3 Relationships between load, shear, and moment diagrams 4 Practical considerations 5 See also 6 References 7 Further reading 8 External links Convention Although these conventions are relative and any convention can be used if stated explicitly, practicing engineers have adopted a standard convention used in design practices. Normal convention The normal convention used in most engineering applications is to label a positive shear force one that spins an element clockwise ( up on the left, and down on the right ). Likewise the normal convention for a positive bending moment is to warp the element in a `` u '' shape manner ( Clockwise on the left, and counterclockwise on the right ). Another way to remember this is if the moment is bending the beam into a `` smile '' then the moment is positive, with compression at the top of the beam and tension on the bottom. Normal positive shear force convention ( left ) and normal bending moment convention ( right ). This convention was selected to simplify the analysis of beams. Since a horizontal member is usually analyzed from left to right and positive in the vertical direction is normally taken to be up, the positive shear convention was chosen to be up from the left, and to make all drawings consistent down from the right. The positive bending convention was chosen such that a positive shear force would tend to create a positive moment. Alternative drawing convention In structural engineering and in particular concrete design the positive moment is drawn on the tension side of the member. This convention puts the positive moment below the beam described above. A convention of placing moment diagram on the tension side allows for frames to be dealt with more easily and clearly. Additionally placing the moment on the tension side of the member shows the general shape of the deformation and indicates on which side of a concrete member rebar should be placed, as concrete is weak in tension. Calculating shear force and bending moment Loaded beam With the loading diagram drawn the next step is to find the value of the shear force and moment at any given point along the element. For a horizontal beam one way to perform this is at any point to `` chop off '' the right end of the beam. The example below includes a point load, a distributed load, and an applied moment. The supports include both hinged supports and a fixed end support. The first drawing shows the beam with the applied forces and displacement constraints. The second drawing is the loading diagram with the reaction values given without the calculations shown or what most people call a free body diagram. The third drawing is the shear force diagram and the fourth drawing is the bending moment diagram. For the bending moment diagram the normal sign convention was used. Below the moment diagram are the stepwise functions for the shear force and bending moment with the functions expanded to show the effects of each load on the shear and bending functions. The example is illustrated using United States customary units. Point loads are expressed in kips ( 1 kip = 1000 lbf = 4.45 kN ), distributed loads are expressed in k / ft ( 1 k / ft = 1 kip / ft = 14.6 kN / m ), moments are expressed in ft - k ( 1 ft - k = 1 ft - kip = 1.356 kNm ), and lengths are in ft ( 1 ft = 0.3048 m ). Step 1 : compute the reaction forces and moments Free - body diagram of whole beam The first step obtaining the bending moment and shear force equations is to determine the reaction forces. This is done using a free body diagram of the entire beam. The beam has three reaction forces, R, R at the two supports and R at the clamped end. The clamped end also has a reaction couple M. These four quantities have to be determined using two equations, the balance of forces in the beam and the balance of moments in the beam. Four unknowns can not be found given two independent equations in these unknown variables and hence the beam is statically indeterminate. One way of solving this problem is to use the principle of linear superposition and break the problem up into the superposition of a number of statically determinate problems. The extra boundary conditions at the supports have to be incorporated into the superposed solution so that the deformation of the entire beam is compatible. From the free - body diagram of the entire beam we have the two balance equations ∑ F = 0, ∑ M A = 0. ( displaystyle sum F = 0 ~, ~ ~ sum M_ ( A ) = 0 ,. ) Summing the forces, we have − 10 − ( 1 ) ( 15 ) + R a + R b + R c = 0 ( displaystyle - 10 - ( 1 ) ( 15 ) + R_ ( a ) + R_ ( b ) + R_ ( c ) = 0 ) and summing the moments around the free end ( A ) we have ( R a ) ( 10 ) + ( R b ) ( 25 ) + ( R c ) ( 50 ) − ( 1 ) ( 15 ) ( 17.5 ) − 50 + M c = 0. ( displaystyle ( R_ ( a ) ) ( 10 ) + ( R_ ( b ) ) ( 25 ) + ( R_ ( c ) ) ( 50 ) - ( 1 ) ( 15 ) ( 17.5 ) - 50 + M_ ( c ) = 0 ,. ) We can solve these equations for R and R in terms of R and M : R b = 37.5 − 1.6 R a + 0.04 M c ( displaystyle R_ ( b ) = 37.5 - 1.6 R_ ( a ) + 0.04 M_ ( c ) ) and R c = − 12.5 + 0.6 R a − 0.04 M c. ( displaystyle R_ ( c ) = - 12.5 + 0.6 R_ ( a ) - 0.04 M_ ( c ) ,. ) If we sum moments about the first support from the left of the beam we have ( 10 ) ( 10 ) − ( 1 ) ( 15 ) ( 7.5 ) + ( R b ) ( 15 ) + ( R c ) ( 40 ) − 50 + M c = 0. ( displaystyle ( 10 ) ( 10 ) - ( 1 ) ( 15 ) ( 7.5 ) + ( R_ ( b ) ) ( 15 ) + ( R_ ( c ) ) ( 40 ) - 50 + M_ ( c ) = 0 ,. ) If we plug in the expressions for R and R we get the trivial identity 0 = 0 which indicates that this equation is not independent of the previous two. Similarly, if we take moments around the second support, we have ( 10 ) ( 25 ) − ( R a ) ( 15 ) + ( 1 ) ( 15 ) ( 7.5 ) + ( R c ) ( 25 ) − 50 + M c = 0. ( displaystyle ( 10 ) ( 25 ) - ( R_ ( a ) ) ( 15 ) + ( 1 ) ( 15 ) ( 7.5 ) + ( R_ ( c ) ) ( 25 ) - 50 + M_ ( c ) = 0 ,. ) Once again we find that this equation is not independent of the first two equations. We could also try to compute moments around the clamped end of the beam to get ( 10 ) ( 50 ) − ( R a ) ( 40 ) − ( R b ) ( 25 ) + ( 1 ) ( 15 ) ( 32.5 ) − 50 + M c = 0. ( displaystyle ( 10 ) ( 50 ) - ( R_ ( a ) ) ( 40 ) - ( R_ ( b ) ) ( 25 ) + ( 1 ) ( 15 ) ( 32.5 ) - 50 + M_ ( c ) = 0 ,. ) This equation also turns out not to be linearly independent from the other two equations. Therefore, the beam is statically indeterminate and we will have to find the bending moments in segments of the beam as functions of R and M. Step 2 : break beam into segments After the reaction forces are found, you then break the beam into pieces. The location and number of external forces on the member determine the number and location of these pieces. The first piece always starts from one end and ends anywhere before the first external force. Step 3 : compute shear forces and moments - first piece Free - body diagram of segment 1 Let V and M be the shear force and bending moment respectively in a cross-section of the first beam segment. As the section of the beam moves towards the point of application of the external force the magnitudes of the shear force and moment may change. This makes the shear force and bending moment a function of the position of cross-section ( in this example x ). By summing the forces along this segment and summing the moments, the equations for the shear force and bending moment are obtained. These equations are : ∑ F = − 10 − V 1 = 0 ( displaystyle sum F = - 10 - V_ ( 1 ) = 0 ) and ∑ M A = − V 1 x + M 1 = 0. ( displaystyle sum M_ ( A ) = - V_ ( 1 ) x + M_ ( 1 ) = 0 ,. ) Therefore, V 1 = − 10 and M 1 = − 10 x. ( displaystyle V_ ( 1 ) = - 10 quad ( \text ( and ) ) quad M_ ( 1 ) = - 10x ,. ) Step 4 : compute shear forces and moments - second piece Free - body diagram of segment 2 Taking the second segment, ending anywhere before the second internal force, we have ∑ F = − 10 + R a − ( 1 ) ( x − 10 ) − V 2 = 0 ( displaystyle sum F = - 10 + R_ ( a ) - ( 1 ) ( x-10 ) - V_ ( 2 ) = 0 ) and ∑ M A = R a ( 10 ) − ( 1 ) ( x − 10 ) ( x + 10 ) 2 − V 2 x + M 2 = 0. ( displaystyle sum M_ ( A ) = R_ ( a ) ( 10 ) - ( 1 ) ( x-10 ) ( \frac ( ( x + 10 ) ) ( 2 ) ) - V_ ( 2 ) x + M_ ( 2 ) = 0 ,. ) Therefore, V 2 = R a − x and M 2 = − 50 + R a ( x − 10 ) − x 2 2. ( displaystyle V_ ( 2 ) = R_ ( a ) - x quad ( \text ( and ) ) quad M_ ( 2 ) = - 50 + R_ ( a ) ( x-10 ) - ( \frac ( x ^ ( 2 ) ) ( 2 ) ) ,. ) Notice that because the shear force is in terms of x, the moment equation is squared. This is due to the fact that the moment is the integral of the shear force. The tricky part of this moment is the distributed force. Since the force changes with the length of the segment, the force will be multiplied by the distance after 10 ft. i.e. ( x-10 ) the moment location is defined in the middle of the distributed force, which is also changing. This is where ( x + 10 ) / 2 is derived from. Alternatively, we can take moments about the cross-section to get ∑ M A = 10 x − R a ( x − 10 ) + ( 1 ) ( x − 10 ) ( x − 10 ) 2 + M 2 = 0. ( displaystyle sum M_ ( A ) = 10x - R_ ( a ) ( x-10 ) + ( 1 ) ( x-10 ) ( \frac ( ( x-10 ) ) ( 2 ) ) + M_ ( 2 ) = 0 ,. ) Again, in this case, M 2 = − 50 + R a ( x − 10 ) − x 2 2. ( displaystyle M_ ( 2 ) = - 50 + R_ ( a ) ( x-10 ) - ( \frac ( x ^ ( 2 ) ) ( 2 ) ) ,. ) Step 5 : compute shear forces and moments - third piece Free - body diagram of segment 3 Taking the third segment, and summing forces, we have − 10 + R a + R b − ( 1 ) ( 15 ) − V 3 = 0 ( displaystyle - 10 + R_ ( a ) + R_ ( b ) - ( 1 ) ( 15 ) - V_ ( 3 ) = 0 ) and summing moments about the cross-section, we get ( 10 ) ( x ) − R a ( x − 10 ) − R b ( x − 25 ) + ( 1 ) ( 15 ) ( x − 17.5 ) + M 3 = 0. ( displaystyle ( 10 ) ( x ) - R_ ( a ) ( x-10 ) - R_ ( b ) ( x-25 ) + ( 1 ) ( 15 ) ( x-17.5 ) + M_ ( 3 ) = 0 ,. ) Therefore, V 3 = 25 − R a − R b = R c ( displaystyle V_ ( 3 ) = 25 - R_ ( a ) - R_ ( b ) = R_ ( c ) ) and M 3 = 262.5 + R a ( x − 10 ) + R b ( x − 25 ) − 25 x = − 675 + R a ( 30 − 0.6 x ) − M c ( 1 − 0.04 x ) + 12.5 x. ( displaystyle M_ ( 3 ) = 262.5 + R_ ( a ) ( x-10 ) + R_ ( b ) ( x-25 ) - 25x = - 675 + R_ ( a ) ( 30 - 0.6 x ) - M_ ( c ) ( 1 - 0.04 x ) + 12.5 x ,. ) Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned. Step 6 : compute shear forces and moments - fourth piece Free - body diagram of segment 4 Taking the fourth and final segment, a balance of forces gives − 10 + R a + R b − ( 1 ) ( 15 ) − V 4 = 0 ( displaystyle - 10 + R_ ( a ) + R_ ( b ) - ( 1 ) ( 15 ) - V_ ( 4 ) = 0 ) and a balance of moments around the cross-section leads to ( 10 ) ( x ) − R a ( x − 10 ) − R b ( x − 25 ) + ( 1 ) ( 15 ) ( x − 17.5 ) − 50 + M 4 = 0. ( displaystyle ( 10 ) ( x ) - R_ ( a ) ( x-10 ) - R_ ( b ) ( x-25 ) + ( 1 ) ( 15 ) ( x-17.5 ) - 50 + M_ ( 4 ) = 0 ,. ) Solving for V and M, we have V 4 = 25 − R a − R b = R c ( displaystyle V_ ( 4 ) = 25 - R_ ( a ) - R_ ( b ) = R_ ( c ) ) and M 4 = 312.5 + R a ( x − 10 ) + R b ( x − 25 ) − 25 x = − 625 + R a ( 30 − 0.6 x ) + M c ( 0.04 x − 1 ) + 12.5 x. ( displaystyle M_ ( 4 ) = 312.5 + R_ ( a ) ( x-10 ) + R_ ( b ) ( x-25 ) - 25x = - 625 + R_ ( a ) ( 30 - 0.6 x ) + M_ ( c ) ( 0.04 x-1 ) + 12.5 x ,. ) By plotting each of these equations on their intended intervals, you get the bending moment and shear force diagrams for this beam. In particular, at the clamped end of the beam, x = 50 and we have M 4 = M c = − 937.5 + 40 R a + 25 R b. ( displaystyle M_ ( 4 ) = M_ ( c ) = - 937.5 + 40R_ ( a ) + 25R_ ( b ) ,. ) Step 7 : compute deflections of the four segments We now use the Euler - Bernoulli beam theory to compute the deflections of the four segments. The differential equation that relates the beam deflection ( w ) to the bending moment ( M ) is d 2 w d x 2 = − M E I ( displaystyle ( \frac ( d ^ ( 2 ) w ) ( dx ^ ( 2 ) ) ) = - ( \frac ( M ) ( EI ) ) ) where E is the Young 's modulus and I is the area moment of inertia of the beam cross-section. Substituting the expressions for M, M, M, M into the beam equation and solving for the deflection gives us w 1 = 5 3 E I x 3 + C 1 + C 2 x w 2 = 1 24 E I x 2 ( x 2 + 600 − 4 R a ( x − 30 ) ) + C 3 + C 4 x w 3 = 1 100 E I ( x 3 3 ( − 625 + 30 R a − 2 M c ) − 50 x 2 ( − 675 + 30 R a − M c ) ) + C 5 + C 6 x w 4 = 1 100 E I ( x 3 3 ( − 625 + 30 R a − 2 M c ) − 50 x 2 ( − 625 + 30 R a − M c ) ) + C 7 + C 8 x ( displaystyle ( \begin ( aligned ) w_ ( 1 ) & = ( \frac ( 5 ) ( 3EI ) ) , x ^ ( 3 ) + C_ ( 1 ) + C_ ( 2 ) , x w_ ( 2 ) & = ( \frac ( 1 ) ( 24EI ) ) , x ^ ( 2 ) , left ( x ^ ( 2 ) + 600 - 4R_ ( a ) ( x-30 ) \right ) + C_ ( 3 ) + C_ ( 4 ) , x w_ ( 3 ) & = ( \frac ( 1 ) ( 100EI ) ) left ( ( \frac ( x ^ ( 3 ) ) ( 3 ) ) ( - 625 + 30R_ ( a ) - 2M_ ( c ) ) - 50x ^ ( 2 ) ( - 675 + 30R_ ( a ) - M_ ( c ) ) \right ) + C_ ( 5 ) + C_ ( 6 ) , x w_ ( 4 ) & = ( \frac ( 1 ) ( 100EI ) ) left ( ( \frac ( x ^ ( 3 ) ) ( 3 ) ) ( - 625 + 30R_ ( a ) - 2M_ ( c ) ) - 50x ^ ( 2 ) ( - 625 + 30R_ ( a ) - M_ ( c ) ) \right ) + C_ ( 7 ) + C_ ( 8 ) , x end ( aligned ) ) ) Step 8 : apply boundary conditions Now we will apply displacement boundary conditions for the four segments to determine the integration constants. For the fourth segment of the beam, we consider the boundary conditions at the clamped end where w = dw / dx = 0 at x = 50. Solving for C and C gives C 7 = − 1250 3 E I ( − 625 + M c + 30 R a ) and C 8 = 125 E I ( − 125 + 6 R a ). ( displaystyle C_ ( 7 ) = - ( \frac ( 1250 ) ( 3EI ) ) ( - 625 + M_ ( c ) + 30R_ ( a ) ) quad ( \text ( and ) ) quad C_ ( 8 ) = ( \frac ( 125 ) ( EI ) ) ( - 125 + 6R_ ( a ) ) ,. ) Therefore, we can express w as w 4 = − 1 300 E I ( x − 50 ) 2 ( − 5 ( 6 R a − 125 ) ( x − 50 ) + 2 M c ( x + 25 ) ). ( displaystyle w_ ( 4 ) = - ( \frac ( 1 ) ( 300EI ) ) ( x-50 ) ^ ( 2 ) left ( - 5 ( 6R_ ( a ) - 125 ) ( x-50 ) + 2M_ ( c ) ( x + 25 ) \right ) ,. ) Now, w = w at x = 37.5 ( the point of application of the external couple ). Also, the slopes of the deflection curves at this point are the same, i.e., dw / dx = dw / dx. Using these boundary conditions and solving for C and C, we get C 5 = − 625 12 E I ( − 5675 + 8 M c + 240 R a ) and C 6 = 250 E I ( 3 R a − 70 ). ( displaystyle C_ ( 5 ) = - ( \frac ( 625 ) ( 12EI ) ) ( - 5675 + 8M_ ( c ) + 240R_ ( a ) ) quad ( \text ( and ) ) quad C_ ( 6 ) = ( \frac ( 250 ) ( EI ) ) left ( 3R_ ( a ) - 70 \right ) ,. ) Substitution of these constants into the expression for w gives us w 3 = 1 300 E I ( 30 R a ( − 50 + x ) 3 − 2 M c ( − 50 + x ) 2 ( 25 + x ) − 625 ( − 141875 + x ( 8400 + ( − 162 + x ) x ) ) ). ( displaystyle ( \begin ( aligned ) w_ ( 3 ) = ( \frac ( 1 ) ( 300EI ) ) ( Bigl ( ) &30R_ ( a ) ( - 50 + x ) ^ ( 3 ) - 2M_ ( c ) ( - 50 + x ) ^ ( 2 ) ( 25 + x ) - &625 ( - 141875 + x ( 8400 + ( - 162 + x ) x ) ) ( Bigr ) ) ,. end ( aligned ) ) ) Similarly, at the support between segments 2 and 3 where x = 25, w = w and dw / dx = dw / dx. Using these and solving for C and C gives C 3 = − 3125 24 E I ( − 1645 + 4 M c + 64 R a ) and C 4 = 25 12 E I ( − 40325 + 6 M c + 120 R a ). ( displaystyle C_ ( 3 ) = - ( \frac ( 3125 ) ( 24EI ) ) ( - 1645 + 4M_ ( c ) + 64R_ ( a ) ) quad ( \text ( and ) ) quad C_ ( 4 ) = ( \frac ( 25 ) ( 12EI ) ) left ( - 40325 + 6M_ ( c ) + 120R_ ( a ) \right ) ,. ) Therefore, w 2 = 1 24 E I ( − 3125 ( − 1645 + 4 M c + 64 R a ) + 50 ( − 4025 + 6 M c + 120 R a ) x + 120 ( 5 + R a ) x 2 − 4 R a x 3 + x 4 ). ( displaystyle ( \begin ( aligned ) w_ ( 2 ) = ( \frac ( 1 ) ( 24EI ) ) ( Bigl ( ) & - 3125 ( - 1645 + 4M_ ( c ) + 64R_ ( a ) ) + &50 ( - 4025 + 6M_ ( c ) + 120R_ ( a ) ) x + 120 ( 5 + R_ ( a ) ) x ^ ( 2 ) - 4R_ ( a ) x ^ ( 3 ) + x ^ ( 4 ) ( Bigr ) ) ,. end ( aligned ) ) ) At the support between segments 1 and 2, x = 10 and w = w and dw / dx = dw / dx. These boundary conditions give us C 1 = − 125 24 E I ( − 40145 + 100 M c + 1632 R a ) and C 2 = 25 4 E I ( − 1315 + 2 M c + 48 R a ). ( displaystyle C_ ( 1 ) = - ( \frac ( 125 ) ( 24EI ) ) ( - 40145 + 100M_ ( c ) + 1632R_ ( a ) ) quad ( \text ( and ) ) quad C_ ( 2 ) = ( \frac ( 25 ) ( 4EI ) ) ( - 1315 + 2M_ ( c ) + 48R_ ( a ) ) ,. ) Therefore, w 1 = 5 24 E I ( 1026125 − 39450 x + 8 x 3 + 20 M c ( − 125 + 3 x ) + 480 R a ( − 85 + 3 x ) ). ( displaystyle w_ ( 1 ) = ( \frac ( 5 ) ( 24EI ) ) left ( 1026125 - 39450x + 8x ^ ( 3 ) + 20M_ ( c ) ( - 125 + 3x ) + 480R_ ( a ) ( - 85 + 3x ) \right ) ,. ) Step 9 : solve for M and R Because w = 0 at x = 25, we can solve for M in terms of R to get M c = 175 − 7.5 R a. ( displaystyle M_ ( c ) = 175 - 7.5 R_ ( a ) ,. ) Also, since w = 0 at x = 10, expressing the deflection in terms of R ( after eliminating M ) and solving for R, gives R a = 25.278 ⟹ M c = − 14.585. ( displaystyle R_ ( a ) = 25.278 quad implies quad M_ ( c ) = - 14.585 ,. ) Step 10 : plot bending moment and shear force diagrams Free - body diagram Shear force diagram Bending moment diagram We can now calculate the reactions R and R, the bending moments M, M, M, M, and the shear forces V, V, V, V. These expressions can then be plotted as a function of length for each segment. Relationship between shear force and bending moment It is important to note the relationship between the two diagrams. The moment diagram is a visual representation of the area under the shear force diagram. That is, the moment is the integral of the shear force. If the shear force is constant over an interval, the moment equation will be in terms of x ( linear ). If the shear force is linear over an interval, the moment equation will be quadratic ( parabolic ). Another note on the shear force diagrams is that they show where external force and moments are applied. With no external forces, the piecewise functions should attach and show no discontinuity. The discontinuities on the graphs are the exact magnitude of either the external force or external moments that are applied. For example, at x = 10 on the shear force diagram, there is a gap between the two equations. This gap goes from - 10 to 15.3. The length of this gap is 25.3, the exact magnitude of the external force at that point. At section 3 on the moment diagram, there is a discontinuity of 50. This is from the applied moment of 50 on the structure. The maximum and minimum values on the graphs represent the max forces and moments that this beam will have under these circumstances. Relationships between load, shear, and moment diagrams Since this method can easily become unnecessarily complicated with relatively simple problems, it can be quite helpful to understand different relations between the loading, shear, and moment diagram. The first of these is the relationship between a distributed load on the loading diagram and the shear diagram. Since a distributed load varies the shear load according to its magnitude it can be derived that the slope of the shear diagram is equal to the magnitude of the distributed load. The relationship between distributed load and shear force magnitude is : d Q d x = − q ( displaystyle ( \frac ( dQ ) ( dx ) ) = - q ) Some direct results of this is that a shear diagram will have a point change in magnitude if a point load is applied to a member, and a linearly varying shear magnitude as a result of a constant distributed load. Similarly it can be shown that the slope of the moment diagram at a given point is equal to the magnitude of the shear diagram at that distance. The relationship between distributed shear force and bending moment is : d M d x = Q ( displaystyle ( \frac ( dM ) ( dx ) ) = Q ) A direct result of this is that at every point the shear diagram crosses zero the moment diagram will have a local maximum or minimum. Also if the shear diagram is zero over a length of the member, the moment diagram will have a constant value over that length. By calculus it can be shown that a point load will lead to a linearly varying moment diagram, and a constant distributed load will lead to a quadratic moment diagram. Practical considerations In practical applications the entire stepwise function is rarely written out. The only parts of the stepwise function that would be written out are the moment equations in a nonlinear portion of the moment diagram ; this occurs whenever a distributed load is applied to the member. For constant portions the value of the shear and / or moment diagram is written right on the diagram, and for linearly varying portions of a member the beginning value, end value, and slope or the portion of the member are all that are required. See also Bending Euler - Bernoulli beam theory Bending moment Singularity function # Example beam calculation References Jump up ^ Livermore C, Schmidt H, Williams J, Socrate S. `` 2.001 Mechanics & Materials I, Fall 2006 ''. Lecture 5 : MIT OpenCourseWare : Massachusetts Institute of Technology. Retrieved 25 October 2013. Jump up ^ `` Moment Diagram Sign Convention Poll ''. Eng - Tips Forum. Retrieved 25 October 2013. Jump up ^ Emweb.unl.edu Jump up ^ Beer, Ferdinand P. ; E. Russell Johnston ; John T. DeWolf ( 2004 ). Mechanics of Materials. McGraw - Hill. pp. 322 -- 323. ISBN 0 - 07 - 298090 - 7. Jump up ^ Hibbeler, R.C ( 1985 ). Structural Analysis. Macmillan. pp. 146 -- 148. Further reading Cheng, Fa - Hwa. `` Shear Forces and Bending Moments in Beams '' Statics and Strength of Materials. New York : Glencoe, McGraw - Hill, 1997. Print. Spotts, Merhyle Franklin, Terry E. Shoup, and Lee Emrey. Hornberger. `` Shear and Bending Moment Diagrams. '' Design of Machine Elements. Upper Saddle River, NJ : Pearson / Prentice Hall, 2004. Print. External links FREE Online Shear Force and Bending Moment Diagram ( SFD & BMD ) Calculator. ( Note : only free up to 3 point loads. ) To draw the shear and moment diagrams by writing the shear and moment equations. To draw the shear and moment diagrams by the relationship between load, shear, and moment. Wikiversity has learning resources about Shear Force and Bending Moment Diagrams Wikimedia Commons has media related to Shear and moment diagram. Retrieved from `` https://en.wikipedia.org/w/index.php?title=Shear_and_moment_diagram&oldid=813263597 '' Categories : Continuum mechanics Structural analysis Diagrams Hidden categories : Commons category without a link on Wikidata Talk Contents About Wikipedia Français Português Română Edit links This page was last edited on 2 December 2017, at 19 : 47 ( UTC ). Text is available under the Creative Commons Attribution - ShareAlike License ; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia ® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. About Wikipedia

Question: what is shear force and bending moment diagram
Answer: analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of a structural element such as a beam