Shear Force and Bending Moment Diagrams V [ N ] x[m] M [ Nm] x[m]
Competencies 1. Draw shear force and bending moment diagrams for point loads and distributed loads 2. Recognize the position of maximum moment as the location of zero shear force. 3. Recognize the moment diagram as the area under the shear force diagram. 2
Review Beam: Concentrated load: Distributed load: A horizontal structural element with vertical loads An external load that is acting on a single point. A external load that is distributed over some area instead of acting on a single point. F L F A F B 3
New External loads cause internal forces. Internal forces cause structural elements to fail. If we can calculate the maximal internal forces, we can look up in a table what structural element is needed (material, cross section area, shape) to withstand those forces. 4
Types of internal forces Tension: Material is pulled apart Cause: Axial loads Compression: Material is squeezed Cause: Axial loads Hint: Think what force would be needed to keep each chunk of the structural element in static equilibrium. 5
Types of internal forces Shear: Material is torn apart vertically Cause: Vertical loads Again: Think what force would be needed to keep each chunk of the structural element in static equilibrium. 6
Types of internal forces Bending Moment: Material is bent Cause: Torques (caused by types of forces) Again: Think what force would be needed to keep each chunk of the structural element in static equilibrium. 7
Shear Force Diagram Shear Diagram: Plotting the shear force V as a function of the distance x from the left side of the beam F L F A V [ N ] x F B x[m] 8
Shear Force Diagram How to draw the shear force diagram: 1. Calculate the reaction forces 1.1 Replace distributed loads by equivalent point loads 1.2 Draw the free body diagram 1.3 Use τ net =0 and F net =0 2. Start at the left with V=0 then move along the beam to the right: - Whenever you encounter an upward vertical load, increase V by the amount of the vertical load. - Whenever you encounter an downward vertical load, decrease V by the amount of the vertical load. 9
Shear Force diagram Example: A beam (negligible mass) is loaded as shown. Draw the Shear Force Diagram (SFD) F L =1kN /m 6.0 m 3.0 m 10.0 m 10
Shear Force Diagram Example continued 0.75 kn F L =1kN /m 2.25 kn 0 m 6 m 9 m 10 m V [kn] Shear Force Diagram 2 1 0 6 9 10 X [m] - 1-2 11
Bending Moment Diagram Bending Moment Diagram: Plotting the bending moment M as a function of the distance x from the left side of the beam F L F A M [ Nm] x F B x[m] 12
Bending Moment Diagram How to draw the bending moment diagram: 1. Draw the shear force diagram (SFD) 2. Start at the left with M=0 then move along the beam to the right. Calculate the total area under the shear force diagram curve from the left: - If the area is above the x-line, count it as positive. - If the area is below the x-line, count it as negative 13
Bending Moment Diagram Our Example: V [kn] 2 Shear Force Diagram 1 0 6 9 10 X [m] - 1-2 M [knm] Bending Moment Diagram 5 4 3 2 1 0 0 6 6.75 9 10 X [m] 14
Bending Moment Diagram Our Example: V [kn] Shear Force Diagram 2 1 0 6 9 10 X [m] - 1-2 V max = -2.25 knm M [knm] Bending Moment Diagram M max is at the location where V=0! M max = 4.78 knm 5 4 3 2 1 0 0 6 6.75 9 10 X [m] 15
Working with Triangular Loads How to solve a problem with a triangular load: F Lmax =21 N /m 1. Don't panic 2. Calculate the equivalent load 3. Calculate the reaction forces F A 14 m F B 4. Calculate some points on the Shear Force Diagram 5. Calculate some points on the Bending Moment Diagram using the equivalent force LEFT OF THE POINT. 16
Working with Triangular Loads How to solve a problem with a triangular load: F Lmax =21 N /m 1. Don't panic 2. Calculate the equivalent load F A 14 m F B 2/3 F L =147 N 17
Working with Triangular Loads How to solve a problem with a triangular load: F L =147 N F Lmax =21 N /m 3. Calculate the reaction forces F A 14 m F B 2/3 F A =49 N F B =98 N 18
Working with Triangular Loads How to solve a problem with a triangular load: F L =147 N F Lmax =21 N /m 4. Calculate some points on the Shear Force Diagram Left end of the diagram: V(x=0m) : Jump from 0 N to + 49 N F A =49 N 14 m F B =98 N Right end of the diagram: V(x=14m) : Jump from -98 N to 0 N x Random point x = 7 Load at x: L(x)=21 N/m / 14 m * x = 10.5 N Total load left of x: LL(x)=1/2*7m*10.5 N = 36.75 N Shear at location x (for x < 14): V(x) = V(0)-LL(x) = 49 N 36.75 N = 12.25 N 19
Working with Triangular Loads How to solve a problem with a triangular load: F L =147 N F Lmax =21 N /m 4. Calculate some points on the Shear Force Diagram Result Excel: F A =49 N x 14 m F B =98 N In case you need every single point Analytical Result: V(x)=49-0.5*21/14*x*x=49-0.75 x 2 20
Working with Triangular Loads How to solve a problem with a triangular load: F L =147 N F Lmax =21 N /m 5. Calculate some points on the Bending Moment Diagram using the equivalent force LEFT OF THE POINT. F A =49 N 14 m F B =98 N Random point x = 7 Total load left of x: LL(x)=1/2*7m*10.5 N = 36.75 N F L =36.75 N x M(x)=Area under V(x): Calculate the area under V(x) at x=7 using an approximated shear force diagram: 49 N 12.25 N V [ N ] M(7m)=[49*2/3+12.25*1/3]N*7m = 257.25 Nm x[m] 21
Working with Triangular Loads How to solve a problem with a triangular load: F L =147 N F Lmax =21 N /m F A =49 N x 14 m F B =98 N 22
If you knew calculus... If you knew calculus Once we found V(x) Here (Page 20): V(x)=49-0.75 x 2 M(x) could be calculated as the integral of V(x) (which is equal to calculating the area under V(x) as: x M (x)= 0 49 0.75 x 2 dx=49 x 0.75 x3 3 23
More Free online beam calculator https://skyciv.com/free-beam-calculator/ Create your own practice problems with solutions. (Or cheat when doing homework) 24