SAN3701 Assignment 01 2026 |Structural Analysis IV |

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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SAN3701: Structural Analysis IV A

Assignment 01 — Semester 1, 2026

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SAN3701
Module Code:
Structural Analysis IV A
Module Name:
Assignment 01 – 2026
Assignment:




Submitted in partial fulfilment of the requirements for SAN3701 – UNISA 2026

,UNISA | SAN3701 Structural Analysis IV A – Assignment 01



Question 1 [30 Marks]: Flexibility Method – Frame Analysis

Question: Determine the support reactions of the frame illustrated in Figure 1 using the
flexibility method, and sketch the shear force and bending moment diagrams for all members.
The flexural rigidity (EI) is constant throughout all members of the structure.

The frame has a fixed support at A, a roller support at D, and is subjected to a uniformly
distributed load (UDL) of 25 kN/m acting horizontally on column AB (height 12 m), and a
200 kN point load acting downward at midspan C of beam BD (total span 12 m, so C is 6 m
from B and 6 m from D).


1.1 Degree of Indeterminacy


The frame has a fixed support at A (providing three reactions: MA , Ax , Ay ) and a roller at D
(providing one vertical reaction: Dy ). The total number of reactions is four (4). For a plane
frame, the equations of static equilibrium available are three (3). Therefore:



Degree of Indeterminacy = 4 − 3 = 1


The frame is statically indeterminate to the first degree.


1.2 Selection of Redundant


Dy is selected as the redundant. Removing Dy produces the primary (released) structure,
which is a cantilever frame fixed at A, free at D.


1.3 Primary Structure Reactions (Real Load, Dy = 0)


With Dy removed, the primary structure carries the full applied loading. Taking moments
about A for the primary structure:

Horizontal reaction at A:

The UDL of 25 kN/m acts over the full height of 12 m on member AB (horizontal load):


Ax = 25 × 12 = 300 kN (← , pointing into the structure)



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,UNISA | SAN3701 Structural Analysis IV A – Assignment 01


Vertical reaction at A:

Taking moments about A (sum of vertical forces, no vertical reaction at D in primary):

X
Fy = 0 : Ay − 200 = 0 =⇒ Ay = 200 kN ↑



Fixed-end moment at A:

Taking moments about A (clockwise positive):

X
MA = 0 :


The UDL on AB acts at a resultant of 25 × 12 = 300 kN at mid-height (h/2 = 6 m above A):


MA = (25 × 12 × 6) + (200 × 6)


MA = 1800 + 1200 = 3000 kN · m


Therefore:
MA = 3000 kN · m, Ax = 300 kN, Ay = 200 kN



1.4 Compatibility Equation


The deflection at D (in the direction of Dy , i.e., vertical) must be zero in the real structure.
The compatibility condition is:



∆D0 + Dy · δDD = 0


where:

• ∆D0 = vertical deflection at D due to applied loads on the primary structure
• δDD = vertical deflection at D due to a unit load Dy = 1 applied at D on the primary
structure


1.5 Computation of ∆D0 (Deflection at D due to Real Loads)




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,UNISA | SAN3701 Structural Analysis IV A – Assignment 01



Bending Moment Diagram for Primary Structure (M0 diagram)


Member AB (column, 0 to 12 m from A):

At distance z from A (upward), the moment due to horizontal UDL is:

25z 2
MAB (z) = MA − Ax · z + = 3000 − 300z + 12.5z 2
2

At z = 0 (A): M = 3000 kN·m; at z = 12 m (B): M = 3000 − 3600 + 1800 = 1200 kN·m.

Member BC (beam, 0 to 6 m from B):

At B, the moment from the column is 1200 kN·m (hogging). Along BC with Ay = 200 kN
acting upward at A (transferred as shear through column), and no horizontal load on beam:


MBC (x) = 1200 − 200x (x measured from B)


At x = 0 (B): M = 1200 kN·m; at x = 6 m (C): M = 1200 − 1200 = 0 kN·m.

Member CD (beam, 0 to 6 m from C toward D):

At C the moment is 0. No loads act on CD in the primary structure (200 kN already applied
at C). At D the moment is 0 (free end in primary). Therefore MCD = 0.


Unit Load Diagram (M̄ diagram, Dy = 1 kN)


Apply Dy = 1 kN upward at D. The primary structure is a cantilever from A.

Reactions due to unit load:


Āy = −1 kN (downward), Āx = 0, M̄A = −12 kN · m (hogging)



Member CD: At distance s from D: M̄CD (s) = 1 × s = s (linear, 0 at D, 6 at C).

Member BC: At distance x from B (x=0 at B, x=6 at C):


M̄BC (x) = −12 + 1 × (12 − x) = −12 + 12 − x = −x


Actually, measuring from B toward C (x from B): The unit upward load at D produces Āy =




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, UNISA | SAN3701 Structural Analysis IV A – Assignment 01


−1 kN. At B: M̄B = M̄A + Āx × 12 = −12 kN·m. Along BC from B:


M̄BC (x) = −12 + 1 × x = x − 12 (x from B)


At x = 0: −12 kN·m; at x = 6: −6 kN·m.

Member AB: No horizontal unit load. M̄AB (z) = −12 kN·m (constant, since Āx = 0).

Wait – correcting for the unit load path: unit load D̄y = 1 kN acts upward at D. Sum mo-
ments about A:
M̄A = 1 × 12 = 12 kN · m (counterclockwise)

Āy = 1 kN (downward, to balance)

Āx = 0


Member BC (from B at x=0 to C at x=6, then CD from C to D):

Cutting at x from B:
M̄BC (x) = M̄A − Āy · x = 12 − 1 · x

At x = 0 (B): M̄ = 12 kN·m; at x = 6 (C): M̄ = 6 kN·m.

Member CD (from C at x=6 to D at x=12, let s = x − 6 from C):


M̄CD (s) = 6 − 1 · s


At s = 0 (C): M̄ = 6 kN·m; at s = 6 (D): M̄ = 0.

Member AB (vertical column, Āx = 0): M̄AB = M̄A = 12 kN·m (constant throughout, since
no horizontal force on column from unit load).


Virtual Work Integral

Z
1
∆D0 = M0 · M̄ ds
EI

Member AB (M0 is parabolic, M̄ = 12 constant, length = 12 m):

Area of M0 diagram over AB:


M0 (0) = 3000, M0 (12) = 1200, shape: parabolic (due to UDL)


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