This article presents a stepwise approach to determining the span moments of a three-span continuous beam, thereby generating its shear force and bending moment diagrams. And at the end of this article, the analysis results from hand computation are compared with that generated using Staad Pro software.

STEPS in Determining the Span Moments and Plotting the BMD

The following steps shall be taken towards determining the span moments:

STEP 1:      Analyze the beam using moment distribution method to obtain the support moments

STEP 2:      Discretize the beam and obtain the reaction at each support using the equations of static equilibrium

STEP 3:      Obtain the Shear force acting on the beam at appropriate section and draw the shear force diagram

STEP 4:      Obtain the Span moment acting on the beam by computing the area under the shear force diagram

STEP 5:      Draw the bending moment diagram.

 

Worked Example:

The continuous span beam to be analyzed is shown below:

 

STEP 1: Analyze the beam using moment distribution method to obtain the support moments.

This article is primarily about determining the span moments of continuous beam and plotting the bending moment diagram. So that the article does not become extensively long, a very detailed explanation of analyzing the support moment for this same continuous beam using moment distribution method has been published here. If you are confident of your ability in using moment distribution method to obtain support moments of indeterminate beams then you should continue with this article. However, if you feel you need some brush ups, I recommend you visit the referenced link (click here) and then return to continue reading this article.

We shall only enumerate the results leading us to evaluating the support moments and not give further explanation in this article.

 

End Moments

$
_{M_{AB\,\,}}\,\,=\,\,-\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{-400\,\,*\,\,\begin{array}{l}
6\\
\end{array}}{8}
$ = -300

$
_{M_{BA\,\,}}\,\,=\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{400\,\,*\,\,\begin{array}{l}
6\\
\end{array}}{8}
$ = 300

$
_{M_{BC\,\,}}\,\,=\,\,-\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{-150\,\,*\,\,\begin{array}{l}
4\\
\end{array}}{8}
$  = -75

$
_{M_{CB\,\,}}\,\,=\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{150\,\,*\,\,\begin{array}{l}
6\\
\end{array}}{8}
$  = 75

$
_{M_{CD\,\,}}\,\,=\,\,-\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{-200\,\,*\,\,\begin{array}{l}
6\\
\end{array}}{8}
$ = -150

$
_{M_{DC\,\,}}\,\,=\,\,\frac{P\,\,*\,\,\begin{array}{l}
L\\
\end{array}}{8}
$ = $\frac{200\,\,*\,\,\begin{array}{l}
6\\
\end{array}}{8}
$  = 150

STIFFNESS

$
_{K_{AB\,\,}}$    = ¾ * 4EI/L =  3EI/6  =  0.5EI

$
_{K_{BC\,\,}}$    =  4EI/L =  4EI/4  – EI

$
_{K_{CD\,\,}}$    = ¾ * 4EI/L =  3EI/6  =  0.5EI

Distribution Factors

$
_{D_{BA\,\,}}\,\,=\,\,\frac{K_{A\underset{}{B}}}{K_{AB}\,\,+\,\,K_{BC}}\,\,=\,\,\,\,\frac{K_{AB}}{K_{Total}}
$ = $\,\,\,\,\frac{0.5EI}{0.5EI\,\,+\,\,EI}\,\,=\,\,0.333
$

$
_{D_{BC\,\,}}\,\,=\,\,\frac{K_{B\underset{}{C}}}{K_{BC}\,\,+\,\,K_{AB}}\,\,=\,\,\,\,\frac{K_{BC}}{K_{Total}}
$ = $\,\,\,\,\frac{EI}{EI\,\,+\,\,0.5EI}\,\,=\,\,0.667
$

$
_{D_{CB\,\,}}\,\,=\,\,\frac{K_{C\underset{}{B}}}{K_{CB}\,\,+\,\,K_{CD}}\,\,=\,\,\,\,\frac{K_{CB}}{K_{Total}}
$ = $\,\,\,\,\frac{EI}{0.5EI\,\,+\,\,EI}\,\,=\,\,0.667
$

$
_{D_{CD\,\,}}\,\,=\,\,\frac{K_{C\underset{}{D}}}{K_{CBB}\,\,+\,\,K_{CD}}\,\,=\,\,\,\,\frac{K_{CD}}{K_{Total}}
$ = $\,\,\,\,\frac{0.5EI}{0.5EI\,\,+\,\,EI}\,\,=\,\,0.333
$

 

Moment Distribution Table

 

STEP 2: Discretize the beam and obtain the reaction at each support

Since the support moments are obtained through moment distribution, we can now obtain the reactions using superposition method. We will calculate the reactions on the beam when they are fixed as well as when they are free (released). We will then find the algebraic sum of the reactions at each point to obtain the actual reactions on the beam.

Estimation of the fixed support reactions

We shall split the beams and make their end fixed without any load acting to get the fixed reactions as shown below:

 

From Beam AB

Taking Moment about A (clockwise direction +)

–R_BA1 x 6 + M_BA – M_AB= 0

R_{BA1 =} (294.8 – 0)/6

R_BA1 = 49.1KN

Summation of vertical forces must be equal to zero

R_{BA1 + RAB1 = 0}

_{RAB1 = -49.1KN}

From Beam BC

Taking Moment about B (clockwise direction +)

–R_CB1 x 4 + M_CB – M_BC = 0

R_{CB1 =} (123.3 – 294.8)/4

R_CB1 = 42.87KN

Summation of vertical forces must be equal to zero

R_{BC1 + RCB1 = 0}

_{RCB1 = -42.87KN}

From Beam CD

Taking Moment about C (clockwise direction +)

–R_DC1 x 6 + M_DC – M_CD = 0

R_{DC1 =} (0 – 123.3)/4

R_DC1 = -20.5KN

Summation of vertical forces must be equal to zero

R_{DC1 + RCD1 = 0}

_{RCD1 = 20.5KN}

Estimation of the free support reactions

From Beam AB

Taking Moment about A (clockwise direction +)

–R_BA2 x 6 + 400 x 3= 0

R_{BA2 =} (-400 x 3)/6

R_BA2 = 200KN

Summation of vertical forces must be equal to zero

R_{BA2 + RAB2 – 400 = 0}

_{RAB2 = 200KN}

From Beam BC

Taking Moment about B (clockwise direction +)

–R_CB2 x 4 + 150 x 2 = 0

R_{CB2 =} (- 150 x 2)/4

R_CB2 = 75KN

Summation of vertical forces must be equal to zero

R_{CB2 + RBC2 – 150 = 0}

_{RBC2 = 75KN}

From Beam CD

Taking Moment about C (clockwise direction +)

–R_DC2 x 6 + 200 x 3 = 0

-R_{DC2 =} (- 200 x 3)/6

R_DC2 = 100KN

Summation of vertical forces must be equal to zero

R_{DC2 + RCD2 – 200 = 0}

_{RCD2 = 100KN}

Actual reactions (summation of all the reactions at each supports)

To get the actual reaction forces, all the reaction forces from the fixed beam and free beam are summed together

R_{A =} –R_{AB1 +} R_{AB2 =} (- 200 x 3)/6 = 49.13 + 200 = 150.9KN

R_{B = (}R_{BA1 +} R_BC1 ) + (R_{BA2 +} R_BC2) = (49.13 + 42.87) + (200 + 75) = 367KN

R_{C = (}R_{CB1 +} R_CD2) + (R_{CB2 +} R_CD2) = (-42.8 + 20.5) + (75 + 100) = 152.7KN

R_{D =} –R_{DC1 +} R_{DC2 =} (- 200.5 + 100) = 79.5KN

Since all the reactions are known, we can now compute the internal shear at desired section and then draw the shear force diagram.

STEP 3:      Obtain the Shear force acting on the beam and draw the shear force diagram

The shear force diagram is obtained by summing up the reactions and forces acting at each section

Shear force at A = 150.9KN

Shear Force under 400KN Load = 150.9 – 400 = -249.1KN

Shear Force at B = -249.1 + 367 = 117.8KN

Shear Force under 150KN Load = 117.8 -150 = -32KN

Shear Force at C = -32 + 152.7 = 120.5

Shear Force under 200KN point Load = 120.5 – 200 = -79.5KN

Shear force at D = -79.5 + 79.5 = 0

These values are plotted to get a shear force diagram as shown below:

STEP 4:      Obtain the Span moment acting on the beam by calculating the change in area under the shear force diagram

 

Span Moment under 400KN Load

Area under shear forced diagram (within the colored region) = 150.8 x 3 = 452.6KNm

Preceding moment (support moment at A) = 0

Span Moment (change in moment) = 453 – 0 = 453KNm

 

Span Moment under 150KN Load

Area under shear forced diagram (within the colored region) = 118 x (8-6) = 236KNm

Preceding moment (support moment at A) = 294.8

Span Moment (change in moment) = 236 – 294.8 = -59KNm

 

Span Moment under 200KN Load

Area under shear forced diagram (within the colored region) = -79.4. x (13-16) = 238KNm

Preceding moment (support moment at D) = 0

Span Moment (change in moment) = 238 – 0 = 238KNm

 

STEP 5: Plot the Bending Moment diagram

Since the Support moments at each support and the span moments under each load has been calculated then we can plot the bending moment diagram as shown.

 

 

The Staad Pro model

The continuous beam is modeled in Staad Pro. The beam is assigned a section geometry of 225 x 450mm and the material is chosen as concrete so that the program can compute the second moment area and determine deformation properties like elastic modulus. A pin supports is assigned to node A, B, C, and D; while 400KN, 150KN and 200KN loads are assigned as nodal loads in the middle of beam AB, BC, and CD respectively.  An analytical model of the beam is shown below.

 

Results and Graphs

 

Shear Force Diagram from Staad Pro

 

Bending Moment diagram from staad Pro

 

 

 

Conclusion

You would observe from the table and also from the graphs that the bending moment and shear force results from hand computation and analysis using Staad Pro software are very close and almost identical. This is to show that this powerful computer programs are also based on structural engineering first principles. Understanding these principles makes a great difference in how an engineer uses these tools (software program), impacts his engineering judgements, and instills confidence in him when taking decisions. Cheers!