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Axial Forces, Shearing Forces and Bending Moment Concept

Axial Forces, Shearing Forces and Bending Moment Diagram For Determinate Frames

  • Determinancy and Indeterminancy of Frame
    The structure is statically determinate if all the external, as well as internal unknown forces, reactions and moments developed in the member, can be completely determined by using equations of static equilibrium and conditional equations,else it is statically indeterminate.
    • Degree of Static Indeterminacy
      D= 3m + r - (3j + c)
    • Degree of External Static Indeterminacy
      DSE = r - (3 + c)
    • Degree of Internal Static Indeterminacy
      DSI = D- DSE
  • Axial Force (AF)
    It is the algebraic sum of all the forces acting along the longitudinal axis of the member on either side of the considered section.
    Figure - 01
  • Shear Force (SF)
    It is the algebraic sum of all the forces on the left-hand side or to the right-hand side (perpendicular to the axis of the member) of the considered section.
    Figure - 02
  • Bending Moment (BM)
    It is the sum of the moment of all the forces on either side of considered section.
    Figure - 03

In order to draw Axial Force(AF), Shearing Force(SF) and Bending Moment(BM) diagrams we follow the following steps:

  • Step 1
    Check whether the structure is statically determinate or not. If yes, then continue for further analysis.
  • Step 2
    Determine the support reactions using condition of static equilibrium and conditional equations if any.
  • Step 3
    Draw free body diagrams (FBD) for each part separating at joints.
  • Step 4
    Calculate AF, SF, and BM for each part from free body diagrams.
  • Step 5
    Draw AF, SF, and BMD for each part.
  • Step 6
    Combine each part with respective diagrams to get the diagram of the whole frame.

Plotting Axial Force, Shear Force, and Bending Moment Diagrams
These Diagrams can be drawn by two methods:

  1. By Mathematical Equation
    To formulate mathematical equation, the method of the section can be used applying equation of equilibrium. AF, SF, and BM calculation become tedious by section method when a number of forces are acting on the beam.
  2. By Calculation of Ordinate
    In this method, equation according to the definition is used. AF, SF, and BM are calculated at small intervals and then plotted in Cartesian coordinate system. After joining all points by the smooth curve, we find AF, SF, and BM. This method is explained with the help of an example below.
    Consider a simply supported beam AB with a concentrated load P
    Figure - 04
    note that H = 0
    \(R_A\) = \(\frac{Pb}{L}\)         \(R_B\) = \(\frac{Pa}{L}\)
    Figure - 05
    for 0 < x < a
    V = \(R_A\) = \(\frac{Pb}{L}\)
    M = \(R_A x\) = \(\frac{Pb x}{L}\)
    note that \(\frac{dM}{dx}\) = \(\frac{Pb}{L}\) = V
    Figure - 06
    for a < x < L
    V = \(R_A - P\) = \(\frac{-Pa}{L}\)
    M = \(R_A x - P(x - a)\) = \(\frac{Pa(L - x)}{L}\)
    note that \(\frac{dM}{dx}\) = \(\frac{-Pa}{L}\) = V
    Figure - 07
    with \(M_{max}\) = \(\frac{Pab}{L}\)
    Figure - 08
    Figure - 09

Super-position Theorem
The principle of Super-position states as "Axial Force, Shear Force, and Bending Moments due to each force may be added to obtain the resulting Axial Force, Shear Force, and Bending Moments when all forces act at once."
The structure must be linearly elastic to use this principle.

  • Importance
    Superposition allows separating the loads in any desired way, analyzing the structure for a separate set of loads and finding the resultant for the sum of loads by adding individual load effects.
  • Method of Superposition
    • Step 1
      Divide given combinations of loads into a number of single loading system for the beam.
    • Step 2
      Calculate support reactions by using basic equations of static equilibrium for all system of loadings.
    • Step 3
      Draw AFD, SFD, and BMD for all system of loadings.
    • Step 4
      Add AFD, SFD, and BMD of all individual system to get resultant AFD, SFD, and BMD for given system of loading.
  • Example:
    Draw the SF and BM Diagram for the Given Beam Using Super-position.
    Figure - 09
    Since no horizontal forces are acting on it so no axial force is developed. But the procedure is same for AF as below.
    Figure - 10
    Figure - 11

Relationship Between Rate of Loading, Shear Force and Bending Moment
Consider the following beam section with a uniformly distributed load with load intensity q. The uniformly distributed loads are positive when they act downward and negative when they act upward.

Figure - 12

Summing forces vertically (+ve) we find
V - qdx - (V + dV) = 0
-qdx - dV = 0
\(\frac{dV}{dx}\) = -q
dV = -qdx
Thus, the shear force varies with x, and the rate of change with respect to x equals to –q. Also, if q = 0, then the shear force is constant.
Integrate between points A and B on the beam we have:
\(\int_A^B dV\) = - \(\int_A^B qdx\)
VB – VA = Area of load intensity diagram between A and B
Summing moments (ccw +ve) we find:
\(M + dM - (M + Vdx) + (qdx) \times \frac{dx}{2}\) = 0
\(dM\) - \(Vdx\) + \(\frac{qdx^2}{2}\) = 0
Discarding products of differentials as they give very small values compared to other terms, we have:
dM - Vdx = 0
dM = Vdx
Integrate between points A and B on the beam we have:
\(\int_A^B dM \) = \(\int_A^B Vdx \)
MB – MA = Area of shear force diagram between A and B

Maximum Shear Force and Bending Moment
The loading on most beams is such that the stress resultant on planes perpendicular to the axis of the beam that consists of a shear force V and a bending moment M. In determining beam responses, it is very convenient and essential to first determine the shear and bending moment diagrams.
The basic procedure to determine the shear and moment diagram is to determine the values of V and M at various locations along the beam and plotting the result.
We determine critical sections within the beam. A critical section is one where a maximum stress occurs.

  • Section of Maximum Shear Force 
    Since the shear force V, at any section of the beam is the algebraic sum of transverse forces to the left of the section i.e. the shear, in most cases, can be found out at a glance.
  • Section of Maximum Moment 
    It can be shown mathematically that if the shear force is zero or changes its sign the bending moment will be either maximum or relative maximum.
    i.e. when \(\frac{dM}{dx}\) = V = 0
    Thus, the bending moment diagram is a constant and continuous curve.
  • Nature of Diagram
Figure - 13
Figure - 14
Figure - 15
Figure - 16
Figure - 17
Figure - 18
Figure - 19
Figure - 20
Figure - 21
Figure - 22
Figure - 23
Figure - 24
Figure - 25
Figure - 26
Figure - 27
Figure - 28
Figure - 29
Figure - 30

Motra, G. B. (2015).Professor, Ph.D. Kathmandu: Heritage Publishers & Distributors Pvt. Ltd.
Bhavikatti, S. S. (1997): 'Strength of Materials' New Delhi: Vikas Publishing House Pvt. Ltd.
Norris, Charles Head / Wilbur, Johan Benson / Utku, Senol (1991): 'Elementary Structural Analysis', McGraw Hill, Singapore.


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