Bending Moment and Shear Forces- Analysis of Structures-Part-4
Bending moment is algebraic sum of all the moments, including moment of support reactions, acting on either side of a section. Similarly, Shear force is algebraic sum of all the forces, including support reactions, acting on either side of a section. In this article you will find answer to the question- How do you calculate shearing forces and bending moments? I will explain it in such a practical way that the people with limited knowledge also can get benefit from this article. Engineering students and those who want to learn structural engineering will get benefit from this article.
After determining the support reactions the bending moments and shear forces in the structure are calculated at different places of the structure. The diagram that shows a variation of Bending moment and Shear force in a structure is called Bending Moment Diagram and Shear Force Diagram respectively.
After knowing the variation of Bending moment and Shear forces the section of the structure is designed. If there is an axial force in the structure that also needs to be determined before designing a section. For designing a column, shown in the following figure, an axial force in the column is required. Similarly, an axial force in the simply supported beam is required too.
Determination of Shear force and Bending Moments: Procedure for determination of Shear force and Bending Moments:
2. Determine if the structure is statically determinate or indeterminate. In the case of indeterminate structures the calculation of support reactions won't be possible by usual method, i.e. using the equilibrium equations. Note: In the upcoming examples the structures will be structurally determinate.
The procedure for determining these diagrams is best described by the calculations shown below.
Step-1 Pre-determination of support reactions: Since the loads are acting vertically downward, the support reactions (Rav and Rbv) must act in vertically upward direction (parallel to axis-z). Horizontal component of P2 causes a horizontal reaction Rah at support-A in the direction parallel with axis-x. Since support B is a roller support the support B cannot produce a horizontal reaction.
Step-2 Determinate or Indeterminate: There are three unknowns, i.e. Rah, Rav and Rbv. We can write three equilibrium equations. Since we have forces acting in axis-z and axis-x we can write Fz=0, Fx=0. Similarly, these forces can produce moment about axis-y, we can write a moment equation, i.e. My=0. Since number of equations are equal to number of unknowns, this structure is a statically determinate structure.
Step -3: Calculation of support reactions:
From Fz=0 we have Rav+Rbv-5*(1+1.5+2)-7-5*sin45=0----eq1. From Fx=0 we have Rah-5*cos45=0----eq2. From My=0 Taking moments of all forces about axis-y at support-A we have Rbv*(1+1.5+2)-7*1-5*sin45*(1+1.5)-5*(1+1.5+2)*(1+1.5+2)/2=0----eq3 Here the bold (last part with each force) parts of the equation are the distance between c.g. of force to support -A. This distance is also called an Arm. Note: When we resolve an inclined force, say P, the component parallel to the base is always P*cos a and the component parallel to the perpendicular is always P*sin a.
Let's rearrange eq3, Rbv*4.5-7-5*0.707*2.5-5*4.5*4.5/2=0 From this Rbv*4.5-66.46=0. Therefore, Rbv=66.46/4.5=14.77 kN
Using eq1, Rav+Rbv-22.5-7-3.53=0 or Rav+Rbv-33.03=0. From here we get, Rav=33.03-Rbv. After substituting value of Rbv in this equation, we get Rav=33.03-14.77=18.26 kN.
From eq2, Rah-5*0.707=0, from this the horizontal reaction, Rah=3.54 kN.
Step-4 Preparation of Diagrams
Shear Force Diagram: S.F. is algebraic summation of all the forces acting perpendicular to longitudinal axis of the structure. This summation is carried out only on one side of the section in consideration.
Let's move from support-B to -A along the beam centre line such a way that the axis-x of the x-z coordinate system coincides with the beam longitudinal axis (centre line). This way axis-z is perpendicular to beam longitudinal axis (Axis-x). Now, all the forces causing S.F. are in parallel to axis-z. Consider a section just next to Support-B. The forces acting in parallel to axis-z (perpendicular to beam centre line) is only Rbv which is 14.77 kN. Therefore, ordinate of S.F. diagram at point "a" is 14.77. Since Rbv is acting upward, it's "+".
As we are moving towards the support-A, now, let's stop at the point just before point load P2. At this section, the algebraic sum of the forces that are in parallel to axis-z is 14.77-5*2= 4.77 kN. Therefore, ordinate of S.F diagram at point "b" is 4.77. When we just cross the P2, the sum will become 14.77-5*2-5*sin45= 1.23 kN. Therefore, c=1.23. Moving on, just before P1, the summation is 14.77-5*sin45-5*(1.5+2) = -6.26 kN. Therefore, d= -6.26. Now, just after P1, the summation is 14.77-5*sin45-5*(1.5+2)-7= -13.26 kN, Therefore, e=-13.26. Just before support-A, f = 14.77-5*sin45-5*(1+1.5+2)-7= -18.26.
Bending Moment: B.M. are calculated in the same way as the S.F. but all the forces are multiplied by their perpendicular distance, i.e. distance from c.g. of the forces to the section in consideration. This distance is called an arm of the moment and measured along axis-x.
Again we move form support-B to support-A with the x-z coordinate system mentioned above. Then we will calculate the B.M. at different sections where we need to know the B.M. We do it by calculating the algebraic sum of all the forces (acting in parallel to axis-z) multiplied by their related arms (distances along axis-x between c.g. of the forces and section in consideration).
To do this, we will multiply all the forces that are in parallel with axis-z by their arms (measured in parallel with axis-x). Ordinate of B.M. diagram at point "a" is 14.77*2-5*2*2/2= 19.54 kNm Since B.M. is a calculated by multiplying a force (kN) by its arm distance (m) its unit is kNm. Here, the Rbv causes anti-clockwise moment at point "a", whereas the UDL causes clockwise moment at the same point. This difference is expressed by the "+" and "-" signs. You can reverse the sign convention if you wish.
Maximum B.M. occurs where S.F. is zero. In this case at section "b" the max. B.M. occurs. In order to compute B.M. at section b we need to find the location where S.F. is zero. Let's say "o" is the point where S.F. is zero. Hence we have two similar triangles, i.e. One between o and d another between o and c. We know the ordinates of S.F. at d and c. Let's say horizontal distance between o and d is "L".
Let's equate the ratios of two triangles.
d/L=c/(1.5-L) from here, d=c*L/(1.5-L) or d*(1.5-L)= c*L or 1.5*d-d*L-c*L=0 or 1.5*d- L*(c+d)=0 from here, 1.5*d=L*(c+d) or L=1.5*d/(c+d). Insert values of the ordinates of S.F. diagram then L=1.5*6.26/(1.23+6.26)=1.25 m (horizontal distance between o and d). With this distance known the horizontal distance between o and c is 1.5-1.25=0.25 m.
Now B.M. at section b or at point o, Mmax= 14.77*(2+0.25)-5*(2+0.25)*(2+0.25)/2-5*sin45*0.25= 19.69 kNm. Similarly, at section c, Mc=14.77*(2+1.5/2)-5*(2+1.5/2)*(2+1.5/2)/2-5*sin45*1.5/2=19.06 kN. Therefore, ordinate of B.M. diagram at point c=19.06. Ordinate of B.M. diagram at point d or Md= 14.77*(2+1.5)-5*(2+1.5)*(2+1.5)/2-5*sin45*1.5=15.77 kNm. B.M. at support A has to be zero as the support does not resit moment about axis-y. Let's check that, MA=14.77*(2+1.5+1)-5*(2+1.5+1)*(2+1.5+1)/2-5*sin45*(1.5+1)-7*1=0.00. This means our entire calculation so far is correct.
Axial Force diagram: In a structure axial force acts along centre line of the support structure. In the example of this beam the axial force acts in the centre line of the beam, i.e. along axis-x of the above mentioned x-z coordinate system.
Again we move form support-B to support-A with the x-z coordinate system mentioned above. We will sum up all the forces that act along the axis-x. We find such forces only after we have passed the inclined force P2. The 5*cos45=3.54 kN is the ordinate of Axial Force diagram at point "a". Further moving towards support-B we don't find any more forces that are in parallel with axis-x. Therefore, the ordinate at point "a" is unchanged and thus ordinate at point "b" is 3.54 kN too. If we had any more of such forces we would have added or subtracted them from 5*cos45 depending on their direction of action.
Step-1 Since both the UDL and Point load are acting downward in parallel to axis-z, the support must react with reaction R in upward direction in parallel to axis-z. The loads causes moments about axis-y at the support. Since moment about axis-y is restrained at the fixed support, a reaction moment, M develops at the fixed support. Since there is no support at other end, no support reactions will develop at other end of the cantilever beam.
Step-2 There are two unknowns, i.e. R and M. We can write two equilibrium equations using the loads, i.e. Fz=0 and My=0. They are equations of forces in axis-z and moment of forces about axis-y. Since number of unknowns are equal to number of equilibrium equations, this structure is a statically determinate structure. Therefore, we can find out the unknowns using only equilibrium equations.
Step-3 From Fz=0, 6+5*3-R=0----eq1 From here, the support reaction, R=21 kN. Similarly, from My=0, 6*3+5*3*3/2-M=0----eq2 From here, M= 40.5 kNm
Step-4 Shear Force Diagram: Always move the x-z coordinate system from unsupported end of cantilever to its fixed support. Thus, ordinate of S.F. diagram at point "a", a= 6 kN, similarly, b= 6+5*3= 21 kN.
Bending Moment Diagram: Again lets move the co-ordinate system in the same way from the unsupported end. Thus ordinate of B.M. diagram at point "a", a= 0. Because arm of moment for the forces at section at "a" is 0. Note, moment =force*arm. Ordinate of B.M. diagram at point "b", b= 6*3+5*3*3/2= 40.5 kN.
Since there are no forces acting in the direction parallel to axis-x, there is no axial force in this structure.
Standard formulae for B.M and S.F. for the standard supports with standard loads are available in most of the structural engineering literatures. These standard supports and loads include simply supported and cantilever beams with UDL, Point load, Triangularly varying loads, etc. Following are some of the examples.
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| Example-Standard formulae for B.M. and S.F. |
In this article we have discussed the following:
What are bending moments?, How do you derive the bending moment equation?, What is a bending moment diagram?, How to find Bending Moment, etc. And the same questions for the Shear forces and Axial forces.
Conclusion, in order to calculate S.F., B.M. and Axial Force in a structure, we need to move the x-z co-ordinate system along centre line of the structure such a way that the axis-x coincides with centre line of the structure. Thus, sum of all the forces acting in parallel to axis-z gives us the S.F. The sum shall be calculated only on one side of the section in consideration. Similarly, Algebraic sum of these forces, acting in axis-z, multiplied by their arms gives us the B.M. Algebraic sum of all the forces acting in axis-x gives us the Axial force. Again, we always need to calculate them only on one side of the section in consideration.
In the next article I will explain practically how we calculate now a days the bending moments, shear force, axial forces and support reactions in the Statically Indeterminate Structures.
Related readings: Loads, Structural Supports
Previous article: Determination of Support Reactions
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