UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
SAN2602: Structural Analysis
Assignment 1 – Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
SAN2602
Module Code:
Structural Analysis
Module Name:
Assignment 1
Assignment:
25 May 2026
Due Date:
Mr. Ngonyama
Examiner:
100
Total Marks:
Submitted in partial fulfilment of the requirements for SAN2602 – UNISA 2026
,UNISA | SAN2602 Structural Analysis – Assignment 1
Question 1: Analysis of a Portal Frame with an Internal Hinge [30 Marks]
The frame shown in Figure 1 consists of two columns fixed at A and D, connected by a hor-
izontal beam BE with an internal hinge at the midpoint. The beam carries a uniformly dis-
tributed load (UDL) of 20 kN/m over its full 8 m span. Horizontal point loads act at various
positions along the columns.
20 kN/m
Hinge
C
25 kN B E
5m
15 kN
10 kN
1m
A D
4m 2m 2m
Figure 1: Figure 1: Portal Frame with Internal Hinge
Page 2 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
1.1 Reactions at Supports A and D [9 Marks]
Question: Given the frame in Figure 1, calculate the reactions at A and D.
Frame Description and Given Data
From Figure 1, the geometry and loading are as follows:
• Left column AB: height = 5 + 1 = 6 m (pin support at A, base is 1 m below beam level)
• Right column DE: height = 2.5 + 2.5 = 5 m (pin support at D)
• Beam BE: total length = 4 + 2 + 2 = 8 m, with an internal hinge at mid-span (4 m from B
and 4 m from E)
• UDL on beam: w = 20 kN/m over the full 8 m span
• Horizontal point load of 25 kN acting rightward at joint B (beam level)
• Horizontal point load of 10 kN acting rightward on the left column, 1 m above A (i.e. 5 m
below beam level)
• Horizontal point load of 15 kN acting leftward on the right column, 2.5 m below beam
level
• Support A: pin (provides Ax and Ay )
• Support D: pin (provides Dx and Dy )
Step 1: Identify Unknowns
There are four unknown reactions:
Ax , Ay , Dx , Dy
The internal hinge provides one additional equation (zero moment at the hinge), making the
structure statically determinate.
Step 2: Global Horizontal Equilibrium
Taking rightward as positive:
X
Fx = 0
Page 3 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
Ax + Dx + 25 + 10 − 15 = 0
Ax + Dx = −20 kN · · · (1)
Step 3: Global Vertical Equilibrium
Total UDL resultant:
W = 20 × 8 = 160 kN
X
Fy = 0
Ay + Dy − 160 = 0
Ay + Dy = 160 kN · · · (2)
Step 4: Hinge Condition – Left Portion
The frame is split at the internal hinge. For the left portion (from A to the hinge), taking mo-
ments about the hinge location, clockwise positive.
Geometry for the left portion:
• The hinge is 4 m horizontally from B (i.e. 4 m from the left column axis)
• The hinge is at beam level, which is 6 m above A
Forces acting on the left portion and their moment arms about the hinge:
• Ay : acts upward at A, horizontal distance to hinge = 4 m ⇒ moment = 4Ay (anti-clockwise,
taken as negative)
• Ax : acts horizontally at A, vertical distance to hinge = 6 m ⇒ moment = 6Ax
• UDL on left half of beam (B to hinge): 20 × 4 = 80 kN, acts at 2 m left of hinge ⇒ mo-
ment = 80 × 2 = 160 kN·m (clockwise, positive)
• 25 kN at B: horizontal load at beam level, no vertical distance from the hinge horizontal
axis ⇒ moment = 0 kN·m
• 10 kN on left column: horizontal load, 5 m below beam level (i.e. 5 m below the hinge) ⇒
moment = 10 × 5 = 50 kN·m (clockwise, positive)
Page 4 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
Applying Mhinge = 0 for the left portion:
P
−4Ay − 6Ax + 160 − 50 = 0
Key Note
The UDL moment (160 kN·m) is clockwise about the hinge (load acts downward, resul-
tant is left of hinge). The 10 kN load acts rightward, and because it is 5 m below the
hinge, it creates a clockwise moment. The reaction Ay (upward) at a horizontal dis-
tance of 4 m creates an anti-clockwise moment; Ax (assumed rightward, but will come
out negative) at 6 m below the hinge creates a moment whose sign depends on assumed
direction.
Re-arranging:
4Ay + 6Ax = 110
2Ay + 3Ax = 55 · · · (3)
Step 5: Hinge Condition – Right Portion
For the right portion (hinge to E and column ED), taking moments about the hinge, clockwise
positive.
Geometry for the right portion:
• The right column D is 4 m horizontally to the right of the hinge
• Support D is 5 m below beam level
Forces and their moments about the hinge:
• Dy : acts upward at D, horizontal distance to hinge = 4 m ⇒ moment = 4Dy (anti-clockwise
= negative, so −4Dy on the left side, or equivalently taken as positive anti-clockwise and
set to zero)
• Dx : acts horizontally at D, vertical distance to hinge = 5 m ⇒ moment = 6Dx (using col-
umn height = 5 m, not 6 m, because D is at a different height)
• UDL on right half of beam (hinge to E): 20 × 4 = 80 kN, acts at 2 m right of hinge ⇒
moment = 80 × 2 = 160 kN·m (anti-clockwise from the right side perspective, so clockwise
when seen from right portion equilibrium = negative sign convention depends on orienta-
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, UNISA | SAN2602 Structural Analysis – Assignment 1
tion; the downward load right of hinge is anti-clockwise about the hinge = −160)
• 15 kN on right column: acts leftward, 2.5 m below beam level ⇒ moment = 15 × 2.5 =
37.5 kN·m
Setting up the right-portion moment equation:
Taking the right portion free body, with rightward and upward forces positive, and summing
moments about the hinge (clockwise positive as before):
X
Mhinge, right = 0
The UDL (80 kN downward, 2 m to the right of the hinge) creates an anti-clockwise moment
about the hinge = −160 kN·m.
Dy (upward, 4 m to the right) creates an anti-clockwise moment = −4Dy , but since it bal-
ances the downward loads, it acts in the opposite sense. Reconsidering with clockwise positive
throughout:
4Dy − 5Dx − 160 − 37.5 = 0
4Dy − 5Dx = 197.5
Dividing by 2:
2Dy − 2.5Dx = 98.75 · · · (4)
Step 6: Solve the System of Equations
From equation (2):
Dy = 160 − Ay · · · (5)
From equation (1):
Dx = −20 − Ax · · · (6)
Page 6 of 20
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
SAN2602: Structural Analysis
Assignment 1 – Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
SAN2602
Module Code:
Structural Analysis
Module Name:
Assignment 1
Assignment:
25 May 2026
Due Date:
Mr. Ngonyama
Examiner:
100
Total Marks:
Submitted in partial fulfilment of the requirements for SAN2602 – UNISA 2026
,UNISA | SAN2602 Structural Analysis – Assignment 1
Question 1: Analysis of a Portal Frame with an Internal Hinge [30 Marks]
The frame shown in Figure 1 consists of two columns fixed at A and D, connected by a hor-
izontal beam BE with an internal hinge at the midpoint. The beam carries a uniformly dis-
tributed load (UDL) of 20 kN/m over its full 8 m span. Horizontal point loads act at various
positions along the columns.
20 kN/m
Hinge
C
25 kN B E
5m
15 kN
10 kN
1m
A D
4m 2m 2m
Figure 1: Figure 1: Portal Frame with Internal Hinge
Page 2 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
1.1 Reactions at Supports A and D [9 Marks]
Question: Given the frame in Figure 1, calculate the reactions at A and D.
Frame Description and Given Data
From Figure 1, the geometry and loading are as follows:
• Left column AB: height = 5 + 1 = 6 m (pin support at A, base is 1 m below beam level)
• Right column DE: height = 2.5 + 2.5 = 5 m (pin support at D)
• Beam BE: total length = 4 + 2 + 2 = 8 m, with an internal hinge at mid-span (4 m from B
and 4 m from E)
• UDL on beam: w = 20 kN/m over the full 8 m span
• Horizontal point load of 25 kN acting rightward at joint B (beam level)
• Horizontal point load of 10 kN acting rightward on the left column, 1 m above A (i.e. 5 m
below beam level)
• Horizontal point load of 15 kN acting leftward on the right column, 2.5 m below beam
level
• Support A: pin (provides Ax and Ay )
• Support D: pin (provides Dx and Dy )
Step 1: Identify Unknowns
There are four unknown reactions:
Ax , Ay , Dx , Dy
The internal hinge provides one additional equation (zero moment at the hinge), making the
structure statically determinate.
Step 2: Global Horizontal Equilibrium
Taking rightward as positive:
X
Fx = 0
Page 3 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
Ax + Dx + 25 + 10 − 15 = 0
Ax + Dx = −20 kN · · · (1)
Step 3: Global Vertical Equilibrium
Total UDL resultant:
W = 20 × 8 = 160 kN
X
Fy = 0
Ay + Dy − 160 = 0
Ay + Dy = 160 kN · · · (2)
Step 4: Hinge Condition – Left Portion
The frame is split at the internal hinge. For the left portion (from A to the hinge), taking mo-
ments about the hinge location, clockwise positive.
Geometry for the left portion:
• The hinge is 4 m horizontally from B (i.e. 4 m from the left column axis)
• The hinge is at beam level, which is 6 m above A
Forces acting on the left portion and their moment arms about the hinge:
• Ay : acts upward at A, horizontal distance to hinge = 4 m ⇒ moment = 4Ay (anti-clockwise,
taken as negative)
• Ax : acts horizontally at A, vertical distance to hinge = 6 m ⇒ moment = 6Ax
• UDL on left half of beam (B to hinge): 20 × 4 = 80 kN, acts at 2 m left of hinge ⇒ mo-
ment = 80 × 2 = 160 kN·m (clockwise, positive)
• 25 kN at B: horizontal load at beam level, no vertical distance from the hinge horizontal
axis ⇒ moment = 0 kN·m
• 10 kN on left column: horizontal load, 5 m below beam level (i.e. 5 m below the hinge) ⇒
moment = 10 × 5 = 50 kN·m (clockwise, positive)
Page 4 of 20
,UNISA | SAN2602 Structural Analysis – Assignment 1
Applying Mhinge = 0 for the left portion:
P
−4Ay − 6Ax + 160 − 50 = 0
Key Note
The UDL moment (160 kN·m) is clockwise about the hinge (load acts downward, resul-
tant is left of hinge). The 10 kN load acts rightward, and because it is 5 m below the
hinge, it creates a clockwise moment. The reaction Ay (upward) at a horizontal dis-
tance of 4 m creates an anti-clockwise moment; Ax (assumed rightward, but will come
out negative) at 6 m below the hinge creates a moment whose sign depends on assumed
direction.
Re-arranging:
4Ay + 6Ax = 110
2Ay + 3Ax = 55 · · · (3)
Step 5: Hinge Condition – Right Portion
For the right portion (hinge to E and column ED), taking moments about the hinge, clockwise
positive.
Geometry for the right portion:
• The right column D is 4 m horizontally to the right of the hinge
• Support D is 5 m below beam level
Forces and their moments about the hinge:
• Dy : acts upward at D, horizontal distance to hinge = 4 m ⇒ moment = 4Dy (anti-clockwise
= negative, so −4Dy on the left side, or equivalently taken as positive anti-clockwise and
set to zero)
• Dx : acts horizontally at D, vertical distance to hinge = 5 m ⇒ moment = 6Dx (using col-
umn height = 5 m, not 6 m, because D is at a different height)
• UDL on right half of beam (hinge to E): 20 × 4 = 80 kN, acts at 2 m right of hinge ⇒
moment = 80 × 2 = 160 kN·m (anti-clockwise from the right side perspective, so clockwise
when seen from right portion equilibrium = negative sign convention depends on orienta-
Page 5 of 20
, UNISA | SAN2602 Structural Analysis – Assignment 1
tion; the downward load right of hinge is anti-clockwise about the hinge = −160)
• 15 kN on right column: acts leftward, 2.5 m below beam level ⇒ moment = 15 × 2.5 =
37.5 kN·m
Setting up the right-portion moment equation:
Taking the right portion free body, with rightward and upward forces positive, and summing
moments about the hinge (clockwise positive as before):
X
Mhinge, right = 0
The UDL (80 kN downward, 2 m to the right of the hinge) creates an anti-clockwise moment
about the hinge = −160 kN·m.
Dy (upward, 4 m to the right) creates an anti-clockwise moment = −4Dy , but since it bal-
ances the downward loads, it acts in the opposite sense. Reconsidering with clockwise positive
throughout:
4Dy − 5Dx − 160 − 37.5 = 0
4Dy − 5Dx = 197.5
Dividing by 2:
2Dy − 2.5Dx = 98.75 · · · (4)
Step 6: Solve the System of Equations
From equation (2):
Dy = 160 − Ay · · · (5)
From equation (1):
Dx = −20 − Ax · · · (6)
Page 6 of 20
