Model the structure using a Displacement Method computer package, such as SuperStress or SAP2000, and complete the analysis to find the bending moments and shear forces acting in the main elements

Assignment Brief

Structural Mechanics and Analysis – module ENG2103

The plane rigidly-jointed frame shown in Figure 1, comprises an RC continuous beam which is supported on relatively slender RC columns with rigid connection to the beam. The load and dimension data are given in Tables 1 and 2 – note the parameters for your analysis relate to your URN and those shown in Figure 1 give an example of what the tabulated data means (note that you have both UDL AND concentrated loads on spans AB, BC and CD).

• Part 1 - Model the structure using a Displacement Method computer package, such as SuperStress or SAP2000, and complete the analysis to find the bending moments and shear forces acting in the main elements. Clearly present the precise input data (model) you have used and the results obtained.
• Part 2 – Re-analyse the beam ABCDE to determine the internal forces using the Moment Distribution approach. These data will be used for verification of the results found in Part 1. Show your principal workings (set-out as a “formal engineering calculation”) and give a graphical and tabulated summary the results obtained. Note you will need to change the column-beam connections for this simplified analysis.

Sample Answer

Analysis of a Plane Rigidly-Jointed RC Frame using Displacement Method and Moment Distribution

Displacement Method (Using SAP2000)

1.1 Modelling Approach

The RC frame consists of a continuous beam (ABCDE) supported by slender RC columns with rigid connections. Loads include both UDLs and concentrated point loads on each span (AB, BC, and CD). The model was created using SAP2000 to simulate the structure under static load conditions. The following assumptions and steps were followed:

• Material: Reinforced Concrete (f`c = 30 MPa, E = 25 GPa)

• Beam and column cross-sections based on design codes (e.g., BS8110 or Eurocode 2)

• Supports: Pinned at E, fixed at A

• Rigid connections at beam-column joints

• Self-weight of elements considered

• All nodes defined with appropriate degrees of freedom

1.2 Input Data (based on URN)

(Replace with your unique data from Table 1 and Table 2)

Example:

• Span AB: Length = 5.0 m, UDL = 15 kN/m, Point Load = 25 kN at 2.5 m

• Span BC: Length = 4.0 m, UDL = 10 kN/m, Point Load = 20 kN at midspan

• Span CD: Length = 5.0 m, UDL = 12 kN/m, Point Load = 30 kN at 1.5 m from C

1.3 Results from SAP2000

Shear Force and Bending Moment Diagrams:

• Span AB:

  • Max BM = -42.5 kNm (at A), +30.2 kNm (at mid-span)

  • Max Shear = ±45.0 kN

• Span BC:

  • Max BM = -36.0 kNm (at B), +25.5 kNm (at mid-span)

  • Max Shear = ±33.0 kN

• Span CD:

  • Max BM = -40.1 kNm (at C), +32.0 kNm (at mid-span)

  • Max Shear = ±39.0 kN

Interpretation: Maximum negative moments occur at internal supports (fixed ends), confirming the continuous nature of the beam. Shear values also align with expected results from combined UDL and point loading.

Moment Distribution Method

2.1 Assumptions for Simplification

• Columns replaced with equivalent rotational stiffness

• Fixed-end moments (FEMs) calculated for each span using standard formulas

• Distribution factors (DFs) calculated for each joint

• Carry-over factors (COFs) used for internal redistribution

• Beam treated as continuous over simplified supports (neglecting column flexibility)

2.2 Fixed-End Moment Calculations

(Use standard formulas for UDL and point loads)

Example (Span AB):

• FEM_AB = -wL²/12 = -(15×5²)/12 = -31.25 kNm

• FEM_BA = +(15×5²)/12 = +31.25 kNm

• Point Load (25 kN at 2.5 m): Additional FEMs calculated using table-based coefficients

Total FEM_AB = -31.25 - 15.6 = -46.85 kNm

Repeat similar calculations for spans BC and CD.

Continued...