UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Semester 1 – 2026
⋄
Module Code: SAN3701
Module Name: Structural Analysis IV A
Assignment No.: 01
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for Structural Analysis IV A (SAN3701)
at the University of South Africa.
,UNISA | SAN3701 Structural Analysis IV A – Assignment 01 – 2026
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. The frame has a fixed sup-
port at A, a roller at D, and is subjected to a uniformly distributed load of 25 kN/m acting
horizontally on the column AB (height 12 m), and a concentrated load of 200 kN applied at
mid-span C of the horizontal beam BCD (total span 12 m, with B at 0 m and D at 12 m, C at
6 m from B).
200 kN
C
B D
6m 6m
2512kN/m
m
A
Figure 1: Frame – Question 1 (scaled for illustration)
1.1 Degree of Static Indeterminacy
The frame has a fixed support at A (providing 3 reactions: Ax , Ay , MA ) and a roller at D
(providing 1 reaction: Dy ). The total number of reactions is 4, and the number of equilibrium
equations for a plane frame is 3.
Degree of indeterminacy = 4 − 3 = 1 (1)
The structure is statically indeterminate to the first degree. The vertical reaction at D,
denoted Dy , is selected as the redundant.
1.2 Primary (Released) Structure
The roller at D is removed. The primary structure is a cantilever frame fixed at A, free at D,
and loaded with the actual external loads (25 kN/m UDL on the column, 200 kN at C).
Page 1 of 29
, UNISA | SAN3701 Structural Analysis IV A – Assignment 01 – 2026
Reactions on the primary structure (released system, Dy = 0)
Taking moments about A for the primary structure:
Horizontal equilibrium:
Ax = w · LAB = 25 × 12 = 300 kN (←) (2)
Vertical equilibrium:
Ay = 200 kN (↑) (3)
Moment at A (taking clockwise positive, UDL acts as horizontal force with resultant at
mid-height, 200 kN acts at 6 m from A horizontally):
12
MA = 25 × 12 × + (200 × 6) = 1800 + 1200 = 3000 kNm (clockwise) (4)
2
1.3 Flexibility Coefficients
The compatibility equation for the flexibility method is:
∆D0 + fDD · Dy = 0 (5)
where ∆D0 is the deflection at D in the primary structure due to the real loads, and fDD is
the deflection at D due to a unit load Dy = 1 kN applied at D in the primary structure.
Deflection ∆D0 at D due to real loads
Using virtual work (unit load method), apply a unit upward load at D on the primary struc-
ture and compute m̄ · M0 /EI ds over all members.
R
Member AB (column, length L1 = 12 m):
The real bending moment M0 in AB due to the UDL and 200 kN at C (measuring z upward
from A):
wz 2 25z 2
M0 (z) = −Ax · z + − MA = −300z + − 3000 (6)
2 2
Page 2 of 29
College of Science, Engineering and Technology
⋄
ASSIGNMENT 01
Semester 1 – 2026
⋄
Module Code: SAN3701
Module Name: Structural Analysis IV A
Assignment No.: 01
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for Structural Analysis IV A (SAN3701)
at the University of South Africa.
,UNISA | SAN3701 Structural Analysis IV A – Assignment 01 – 2026
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. The frame has a fixed sup-
port at A, a roller at D, and is subjected to a uniformly distributed load of 25 kN/m acting
horizontally on the column AB (height 12 m), and a concentrated load of 200 kN applied at
mid-span C of the horizontal beam BCD (total span 12 m, with B at 0 m and D at 12 m, C at
6 m from B).
200 kN
C
B D
6m 6m
2512kN/m
m
A
Figure 1: Frame – Question 1 (scaled for illustration)
1.1 Degree of Static Indeterminacy
The frame has a fixed support at A (providing 3 reactions: Ax , Ay , MA ) and a roller at D
(providing 1 reaction: Dy ). The total number of reactions is 4, and the number of equilibrium
equations for a plane frame is 3.
Degree of indeterminacy = 4 − 3 = 1 (1)
The structure is statically indeterminate to the first degree. The vertical reaction at D,
denoted Dy , is selected as the redundant.
1.2 Primary (Released) Structure
The roller at D is removed. The primary structure is a cantilever frame fixed at A, free at D,
and loaded with the actual external loads (25 kN/m UDL on the column, 200 kN at C).
Page 1 of 29
, UNISA | SAN3701 Structural Analysis IV A – Assignment 01 – 2026
Reactions on the primary structure (released system, Dy = 0)
Taking moments about A for the primary structure:
Horizontal equilibrium:
Ax = w · LAB = 25 × 12 = 300 kN (←) (2)
Vertical equilibrium:
Ay = 200 kN (↑) (3)
Moment at A (taking clockwise positive, UDL acts as horizontal force with resultant at
mid-height, 200 kN acts at 6 m from A horizontally):
12
MA = 25 × 12 × + (200 × 6) = 1800 + 1200 = 3000 kNm (clockwise) (4)
2
1.3 Flexibility Coefficients
The compatibility equation for the flexibility method is:
∆D0 + fDD · Dy = 0 (5)
where ∆D0 is the deflection at D in the primary structure due to the real loads, and fDD is
the deflection at D due to a unit load Dy = 1 kN applied at D in the primary structure.
Deflection ∆D0 at D due to real loads
Using virtual work (unit load method), apply a unit upward load at D on the primary struc-
ture and compute m̄ · M0 /EI ds over all members.
R
Member AB (column, length L1 = 12 m):
The real bending moment M0 in AB due to the UDL and 200 kN at C (measuring z upward
from A):
wz 2 25z 2
M0 (z) = −Ax · z + − MA = −300z + − 3000 (6)
2 2
Page 2 of 29
