UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
SAN2602 – ASSIGNMENT 1
Semester 1 Assessment 1 – May 2026
⋄
Module Code: SAN2602
Module Name: Structural Analysis
Assignment No.: Assignment 1
Due Date: 25 May 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for Structural Analysis
at the University of South Africa.
,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 1 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
1.1 Reactions at A and D (9 Marks)
Given information from Figure 1
• UDL on beam BE = 20 kN/m over 8 m total span
• Horizontal load of 25 kN acting rightward at B (beam level)
• Horizontal load of 10 kN acting rightward at 1 m above A
• Horizontal load of 15 kN acting leftward at 2.5 m below E (i.e. 2.5 m from top of right
column)
• Internal hinge at midpoint of beam, 4 m from B and 4 m from E
• Left column height = 5 m (A to B); right column height = 5 m (D to E)
• Support A: pin (two unknowns Ax , Ay ); Support D: pin (two unknowns Dx , Dy )
Step 1: Identify the unknowns
Unknowns: Ax , Ay , Dx , Dy
The internal hinge provides one additional equation (moment of one side about the hinge
equals zero), making the structure statically determinate with four equations for four un-
knowns.
Step 2: Total UDL resultant
Wtotal = 20 × 8 = 160 kN ↓ acting at 4 m from B
Step 3: Horizontal equilibrium of whole frame
Taking rightward as positive:
Page 2 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
X
Fx = 0
Ax + Dx + 25 + 10 − 15 = 0
Ax + Dx + 20 = 0
∴ Ax + Dx = −20 kN (1)
Step 4: Vertical equilibrium of whole frame
Taking upward as positive:
X
Fy = 0
Ay + Dy − 160 = 0
∴ Ay + Dy = 160 kN (2)
Step 5: Hinge condition – Left section (B to hinge)
Cut the frame at the internal hinge. For the left portion, take moments about the hinge loca-
tion. The hinge is 4 m horizontally from B and 5 m vertically above A.
Forces acting on the left portion:
• Ax (horizontal reaction at A), Ay (vertical reaction at A)
• UDL on left half of beam: 20 × 4 = 80 kN, acting at 2 m from B, i.e. 2 m to the left of the
hinge
• 25 kN horizontal at B (at beam level, same height as hinge – no moment arm vertically)
• 10 kN horizontal at 1 m above A, i.e. the hinge is 5 − 1 = 4 m above the 10 kN load
Critical Step
The hinge is at beam level (5 m above ground). The 25 kN load acts at the same height
as the hinge (at B, beam level). Therefore the 25 kN load produces zero moment about
the hinge because the vertical distance between them is zero.
Taking moments about the hinge (clockwise positive), for the left section:
Page 3 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
X
Mhinge, left = 0
Ay · (4) − Ax · (5) − 80 · (2) + 10 · (4) = 0
Note on signs: Ay acts upward at 4 m left of hinge (clockwise +), Ax acts in some direction –
if it acts leftward it creates a moment; taking the sign carefully with Ax measured as positive
rightward:
4Ay − 5Ax − 160 + 40 = 0
4Ay − 5Ax = 120
∴ 4Ay − 5Ax = 120 (3)
Step 6: Hinge condition – Right section (hinge to E)
For the right portion, take moments about the hinge. Forces acting on the right portion:
• Dx , Dy at support D
• UDL on right half of beam: 20 × 4 = 80 kN, acting at 2 m to the right of the hinge
• 15 kN horizontal load pointing left, located 2.5 m below E, i.e. at 5 − 2.5 = 2.5 m above
D. Vertical distance from hinge to this load = 5 − 2.5 = 2.5 m (load is 2.5 m below beam
level)
Taking moments about the hinge (clockwise positive), for the right section:
X
Mhinge, right = 0
−Dy · (4) + Dx · (5) + 80 · (2) + 15 · (2.5) = 0
−4Dy + 5Dx + 160 + 37.5 = 0
−4Dy + 5Dx = −197.5
∴ 4Dy − 5Dx = 197.5 (4)
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, UNISA | SAN2602 Structural Analysis – Assignment 1
Step 7: Solve the four equations simultaneously
From equation (2):
Dy = 160 − Ay (5)
From equation (1):
Dx = −20 − Ax (6)
Substitute (5) and (6) into equation (4):
4(160 − Ay ) − 5(−20 − Ax ) = 197.5
640 − 4Ay + 100 + 5Ax = 197.5
740 − 4Ay + 5Ax = 197.5
−4Ay + 5Ax = 197.5 − 740
−4Ay + 5Ax = −542.5 (7)
Now solve equations (3) and (7) simultaneously. Multiply equation (3) by 1:
4Ay − 5Ax = 120 (3)
−4Ay + 5Ax = −542.5 (7)
Add (3) and (7):
(4Ay − 5Ax ) + (−4Ay + 5Ax ) = 120 + (−542.5)
0 = −422.5
Page 5 of 35
College of Science, Engineering and Technology
⋄
SAN2602 – ASSIGNMENT 1
Semester 1 Assessment 1 – May 2026
⋄
Module Code: SAN2602
Module Name: Structural Analysis
Assignment No.: Assignment 1
Due Date: 25 May 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for Structural Analysis
at the University of South Africa.
,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 1 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
1.1 Reactions at A and D (9 Marks)
Given information from Figure 1
• UDL on beam BE = 20 kN/m over 8 m total span
• Horizontal load of 25 kN acting rightward at B (beam level)
• Horizontal load of 10 kN acting rightward at 1 m above A
• Horizontal load of 15 kN acting leftward at 2.5 m below E (i.e. 2.5 m from top of right
column)
• Internal hinge at midpoint of beam, 4 m from B and 4 m from E
• Left column height = 5 m (A to B); right column height = 5 m (D to E)
• Support A: pin (two unknowns Ax , Ay ); Support D: pin (two unknowns Dx , Dy )
Step 1: Identify the unknowns
Unknowns: Ax , Ay , Dx , Dy
The internal hinge provides one additional equation (moment of one side about the hinge
equals zero), making the structure statically determinate with four equations for four un-
knowns.
Step 2: Total UDL resultant
Wtotal = 20 × 8 = 160 kN ↓ acting at 4 m from B
Step 3: Horizontal equilibrium of whole frame
Taking rightward as positive:
Page 2 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
X
Fx = 0
Ax + Dx + 25 + 10 − 15 = 0
Ax + Dx + 20 = 0
∴ Ax + Dx = −20 kN (1)
Step 4: Vertical equilibrium of whole frame
Taking upward as positive:
X
Fy = 0
Ay + Dy − 160 = 0
∴ Ay + Dy = 160 kN (2)
Step 5: Hinge condition – Left section (B to hinge)
Cut the frame at the internal hinge. For the left portion, take moments about the hinge loca-
tion. The hinge is 4 m horizontally from B and 5 m vertically above A.
Forces acting on the left portion:
• Ax (horizontal reaction at A), Ay (vertical reaction at A)
• UDL on left half of beam: 20 × 4 = 80 kN, acting at 2 m from B, i.e. 2 m to the left of the
hinge
• 25 kN horizontal at B (at beam level, same height as hinge – no moment arm vertically)
• 10 kN horizontal at 1 m above A, i.e. the hinge is 5 − 1 = 4 m above the 10 kN load
Critical Step
The hinge is at beam level (5 m above ground). The 25 kN load acts at the same height
as the hinge (at B, beam level). Therefore the 25 kN load produces zero moment about
the hinge because the vertical distance between them is zero.
Taking moments about the hinge (clockwise positive), for the left section:
Page 3 of 35
,UNISA | SAN2602 Structural Analysis – Assignment 1
X
Mhinge, left = 0
Ay · (4) − Ax · (5) − 80 · (2) + 10 · (4) = 0
Note on signs: Ay acts upward at 4 m left of hinge (clockwise +), Ax acts in some direction –
if it acts leftward it creates a moment; taking the sign carefully with Ax measured as positive
rightward:
4Ay − 5Ax − 160 + 40 = 0
4Ay − 5Ax = 120
∴ 4Ay − 5Ax = 120 (3)
Step 6: Hinge condition – Right section (hinge to E)
For the right portion, take moments about the hinge. Forces acting on the right portion:
• Dx , Dy at support D
• UDL on right half of beam: 20 × 4 = 80 kN, acting at 2 m to the right of the hinge
• 15 kN horizontal load pointing left, located 2.5 m below E, i.e. at 5 − 2.5 = 2.5 m above
D. Vertical distance from hinge to this load = 5 − 2.5 = 2.5 m (load is 2.5 m below beam
level)
Taking moments about the hinge (clockwise positive), for the right section:
X
Mhinge, right = 0
−Dy · (4) + Dx · (5) + 80 · (2) + 15 · (2.5) = 0
−4Dy + 5Dx + 160 + 37.5 = 0
−4Dy + 5Dx = −197.5
∴ 4Dy − 5Dx = 197.5 (4)
Page 4 of 35
, UNISA | SAN2602 Structural Analysis – Assignment 1
Step 7: Solve the four equations simultaneously
From equation (2):
Dy = 160 − Ay (5)
From equation (1):
Dx = −20 − Ax (6)
Substitute (5) and (6) into equation (4):
4(160 − Ay ) − 5(−20 − Ax ) = 197.5
640 − 4Ay + 100 + 5Ax = 197.5
740 − 4Ay + 5Ax = 197.5
−4Ay + 5Ax = 197.5 − 740
−4Ay + 5Ax = −542.5 (7)
Now solve equations (3) and (7) simultaneously. Multiply equation (3) by 1:
4Ay − 5Ax = 120 (3)
−4Ay + 5Ax = −542.5 (7)
Add (3) and (7):
(4Ay − 5Ax ) + (−4Ay + 5Ax ) = 120 + (−542.5)
0 = −422.5
Page 5 of 35
