CE 9: Structural Theory

ANALYSIS OF STATICALLY
DETERMINATE STRUCTURES
JOHN PATRICK L. RONTOS
Instructor, College of Engineering
Samar State University
DETERMINATE STRUCTURES

▪ stable structures
▪ can be analyzed by applying the three (3) basic conditions of equilibrium
✓ σ FH = 0
✓ σ FV = 0
✓ σM = 0
DETERMINATE STRUCTURES (cont’d)

▪ beams
▪ plane trusses
▪ cables and arches
▪ plane frames
STATICALLY
DETERMINATE BEAMS
Procedure for Analysis. Examples.
BEAM

▪ straight member that is loaded perpendicular to its longitudinal axis

Types of beam. Retrieved from: https://learnmech.com/types-of-beam-

classification-of-bea/
PROCEDURE FOR ANALYSIS

Draw the free-body diagram and evaluate the static determinacy of the given
1 beam.

Determine the unknown reactions by applying the three (3) equations of

2 equilibrium. Consider also the presence of internal supports.

Check the computed reactions.

3

Draw the shear and moment diagrams.

4
EXAMPLE 1

Determine the reactions at the supports for the beam shown below. Draw the
shear and moment diagrams.
EXAMPLE 1 (SOLUTION)
▪ Draw the FBD. Assume the direction of the unknown reactions as shown below.

DH
A B C D
MD
DV

▪ Evaluate the static determinacy of the beam. r = 3

r?3
3=3 (statically determinate externally)
EXAMPLE 1 (SOLUTION cont’d)
▪ Apply the equations of equilibrium to calculate the unknown reactions

DH
A B C D
MD
DV

Assume forces directed to the right to be positive.

σ FH = 0
𝐃𝐇 = 𝟎
EXAMPLE 1 (SOLUTION cont’d)
▪ Apply the equations of equilibrium to calculate the unknown reactions

DH = 0
A B C D
MD
DV

Assume forces directed upward to be positive.

σ FV = 0
−15 6 − 160 + DV = 0
𝐃𝐕 = 𝟐𝟓𝟎 𝐤𝐍 (upward)
EXAMPLE 1 (SOLUTION cont’d)
▪ Apply the equations of equilibrium to calculate the unknown reactions

DH = 0
A B C D MD
DV = 250 kN

Assume clockwise moments to be positive.

σ MD = 0

400 − 15 6 3 + 8 − 160 4 − MD = 0

MD = −1,230 kN − m

𝐌𝐃 = 𝟏, 𝟐𝟑𝟎 𝐤𝐍 − 𝐦 (clockwise)
 ▪ Draw the shear and moment diagrams.

MD = 1,230 kN-m

DH = 0
A B C D

DV = 250 kN

Shear Diagram -270 -360

-1,000
-90
2˚ -250
400

130

Moment Diagram
-230

-1,230
EXAMPLE 2

Determine the reactions at the supports for the beam shown below. Draw the
shear and moment diagrams.
EXAMPLE 2 (SOLUTION)
▪ Draw the FBD. Assume the direction of the unknown reactions as shown below.

A B C D E F
AH

AV CV DV FV
▪ Evaluate the static determinacy of the beam. r = 5; ec = 2
r ? 3 + ec
5?3+2
5 = 5 (statically determinate externally)
EXAMPLE 2 (SOLUTION)
▪ Determine the horizontal reaction at A.

A B C D E F
AH

AV CV DV FV

Assume forces directed to the right to be positive.

σ FH = 0
𝐀𝐇 = 𝟎
EXAMPLE 2 (SOLUTION)
▪ Separate the beam into three (3) parts.

1 2 3

A B C D E F
AH = 0

AV CV DV FV
 EXAMPLE 2 (SOLUTION)
▪ Consider section 1. Determine the unknown reactions and internal forces.
Assume the direction of the unknown forces as shown in the FBD of the section.

5 kN/m Assume forces directed to the right to be positive.

BH σ FH = 0
A B
AH = 0
0 − BH = 0
BV BH = 0
AV
Assume clockwise moments to be positive. Assume forces directed upward to be positive.
σ MB = 0 σ FV = 0

AV 20 − 5(20)(10) = 0 50 − 5 20 + BV = 0
𝐀 𝐕 = 𝟓𝟎 𝐤𝐍 (upward) BV = 50 kN (upward)
EXAMPLE 2 (SOLUTION)
SECTION 1

5 kN/m

A B BH = 0
AH = 0

BV = 50 kN
AV = 50 kN
 EXAMPLE 2 (SOLUTION)
▪ Consider section 3. Determine the unknown reactions and internal forces.
Assume the direction of the unknown forces as shown in the FBD of the section.
Assume forces directed to the right to be positive.
3 kN/m
EH E σ FH = 0
F
EH = 0

EV
FV
Assume clockwise moments to be positive. Assume forces directed upward to be positive.
σ ME = 0 σ FV = 0

−FV 20 + 3 20 10 = 0 30 − 3 20 + EV = 0
𝐅𝐕 = 𝟑𝟎 𝐤𝐍 (upward) EV = 30 kN (upward)
EXAMPLE 2 (SOLUTION)
SECTION 3

3 kN/m
EH = 0 E F

EV = 30 kN FV = 30 kN
 EXAMPLE 2 (SOLUTION)
▪ Transfer the internal forces from Sections 1 and 3 to Section 2.

5 kN/m 5 kN/m
3 kN/m
A B BH = 0 B C D E EH = 0 E
AH = 0 F

EV = 30 kN
AV = 50 kN BV = 50 kN BV = 50 kN CV DV EV = 30 kN FV = 30 kN
 EXAMPLE 2 (SOLUTION)
▪ Consider section 2. Determine the unknown reactions. Assume the direction of
the unknown forces as shown in the FBD of the section.
Assume clockwise moments to be positive.
σ MD = 0

5 kN/m −50 70 + CV 50 + 30 20 −
5 20 10 + 50 − 3 70 15 = 0
BH = 0 B C D E EH = 0
𝐂𝐕 = 𝟐𝟒𝟏 𝐤𝐍 (upward)
Assume forces directed upward to be positive.
EV = 30 kN
BV = 50 kN CV DV σ FV = 0

−50 + 241 − 30 − 5 20 − 3 70 + DV = 0
𝐃𝐕 = 𝟏𝟒𝟗 𝐤𝐍 (upward)
 ▪ Draw the shear and moment diagrams.

A B C D E F
AH = 0

AV = 50 kN CV = 241 kN DV = 149 kN FV = 30 kN
x1 x2 x3 50 50 + 150
=
x1 20 + 20
91 90
50
1,200 30 x1 = 10 m
1,380.1667
Shear Diagram 250
-580.1667
150
-150
-30
-2,250
-50 91 91 + 59
-59 =
-150
x2 50
x2 = 30.3333 m

90 90 + 30
250 =
2˚
150 x3 20 + 20
Moment Diagram x3 = 30 m
2˚ 2˚
-619.8333

-1,200
-2,000
STATICALLY
DETERMINATE TRUSSES
Procedure for Analysis. Method of Joints. Method
of Sections. Examples.
TRUSS
▪ a structure composed of slender members joined together at their end points
▪ members are subjected to either axial tension or compression

Common types of bridge truss (1st and 2nd columns) and roof truss (3rd and 4th columns).
TENSION VS. COMPRESSION

Tension

Compression
ASSUMPTIONS

1. All members are connected only at their ends by frictionless hinges in plane
trusses.
2. All loads and support reactions are applied only at the joints.
3. The centroidal axis of each member coincides with the line connecting the
centers of the adjacent joints.
PROCEDURE FOR ANALYSIS

Draw the free-body diagram and evaluate the static determinacy of the given
1 plane truss.

Considering the entire truss, calculate the reactions at the supports using the
2 three (3) basic conditions of equilibrium.

Determine the unknown internal forces in the members by applying either

3 the method of joints or method of sections.

Check the computed internal forces by checking if the joints are in

4 equilibrium.
METHOD OF JOINTS

Determine the slopes of the inclined members of the truss.

1
Examine the FBD of the entire truss to select a joint that has no more than
two unknown forces acting on it. Draw the FBD of the selected joint (including
2
the external forces acting on that joint) and assume the sense or direction of
the unknown forces.
Determine the unknown forces by applying the two equilibrium equations:
3 σ FH = 0 and σ FV = 0.

If all the desired member forces in the selected joint have been determined,
proceed to the next joint. Take that note that the sense of the forces of
4 members connected to the next joint must carried over. Repeat Step 3. Do
this process to the other joints in the truss.
EXAMPLE 3

Determine the force in each member of the Warren truss shown below by the
method of joints.
EXAMPLE 3 (SOLUTION)
▪ Draw the FBD. Assume the direction of the unknown reactions as shown below.

AH

AV EV

▪ Evaluate the static determinacy of the beam. r = 3; m = 13; j = 8

m + r ? 2j
13 + 3 ? 2(8)
16 = 16 (statically determinate)
EXAMPLE 3 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown
below.

AH

AV EV
Assume forces directed to the right to be positive.
σ FH = 0
𝐀𝐇 = 𝟎
EXAMPLE 3 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown below.

AH = 0

AV EV
Assume clockwise moments to be positive.
σ MA = 0
100 6 + 125 12 + 50 18 − EV 24 = 0
EV = 125 kN (upward)
 EXAMPLE 3 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown below.

AH = 0

AV EV = 125 kN
Assume forces directed upward to be positive.
σ FV = 0
AV − 100 − 125 − 50 + 125 = 0
AV = 150 kN (upward)
 EXAMPLE 3 (SOLUTION cont’d)
▪ Determine the slopes of the inclined members.

5 4
3 5
3 3
4 5
4

AH = 0

AV = 150 kN EV = 125 kN
EXAMPLE 3 (SOLUTION cont’d) 3
5 5
3
4
3
4 5
4

▪ Apply method of joints. Consider Joint A. AH = 0

Assume FAF and FAB as tension.
AV = 150 kN EV = 125 kN
Assume forces directed upward to be positive.
σ FV = 0
FAF
3
150 + FAF (5) = 0
FAB
AH = 0 FAF = −250 kN; 𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 (𝐂)

Assume forces directed to the right to be positive.

AV =150
σ FH = 0
4
0 + FAB − 250(5) = 0

𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)

EXAMPLE 3 (SOLUTION cont’d) 3
5 5
3
4
3
4 5
4

▪ Apply method of joints. Consider Joint B. AH = 0

Assume FBF and FBC as tension.
AV = 150 kN EV = 125 kN
Assume forces directed upward to be positive.
FBF
σ FV = 0 𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
−100 + FBF = 0
200 FBC
𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 (𝐓)

Assume forces directed to the right to be positive.

σ FH = 0

−200 + FBC = 0

𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)

EXAMPLE 3 (SOLUTION cont’d) 3
5 5
3
4
3
4 5
4

▪ Apply method of joints. Consider Joint F. AH = 0

Assume FFG and FFC as tension.
AV = 150 kN EV = 125 kN
Assume forces directed upward to be positive.

FFG σ FV = 0 𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂

3 3
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
250 FFC −100 + 250 − FFC (5) = 0 𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
5
𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
100 𝐅𝐅𝐂 =
𝟐𝟓𝟎
𝐨𝐫 𝟖𝟑. 𝟑𝟑𝟑𝟑 𝐤𝐍 (𝐓)
𝟑
Assume forces directed to the right to be positive.
σ FH = 0
4 𝟐𝟓𝟎 4
250 5
+ 𝟑 5
+ FFG = 0
800 𝟖𝟎𝟎
FFG = − or − 266.6667 kN; 𝐅𝐅𝐆 = 𝐨𝐫 𝟐𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐂)
3 𝟑
EXAMPLE 3 (SOLUTION cont’d) 3
5 5
3
4
3
4 5
4

▪ Apply method of joints. Consider Joint G. AH = 0

Assume FGH and FGC as tension.
AV = 150 kN EV = 125 kN
Assume forces directed upward to be positive. 𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
FGH σ FV = 0
𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
800/3 𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
− FGC = 0 𝟐𝟓𝟎
𝐅𝐅𝐂 = 𝐤𝐍 𝐓
𝐅𝐆𝐂 = 𝟎 𝟑
𝟖𝟎𝟎
FGC 𝐅𝐅𝐆 = 𝐤𝐍 (𝐂)
𝟑
Assume forces directed to the right to be positive.
σ FH = 0
𝟖𝟎𝟎
𝟑
+ FGH = 0
800 𝟖𝟎𝟎
FGH = − or − 266.6667 kN; 𝐅𝐆𝐇 = 𝐨𝐫 𝟐𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐂)
3 𝟑
 5 4
3 5 3
3

EXAMPLE 3 (SOLUTION cont’d) 4 4

5

AH = 0

▪ Apply method of joints. Consider Joint C. AV = 150 kN EV = 125 kN

Assume FCH and FCD as tension.
𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
Assume forces directed upward to be positive. 𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
σ FV = 0
𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
250 3 3 𝟐𝟓𝟎
250/3 FCH −125 + + FCH =0 𝐅𝐅𝐂 = 𝐤𝐍 𝐓
3 5 5 𝟑
𝟖𝟎𝟎
𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓) 𝐅𝐅𝐆 = 𝐤𝐍 𝐂
200 FCD 𝟑
𝐅𝐆𝐂 = 𝟎
Assume forces directed to the right to be positive. 𝟖𝟎𝟎
𝐅𝐆𝐇 = 𝐤𝐍 (𝐂)
𝟑
σ FH = 0
250 4 4
−200 − ( )+ 125(5) + FCD = 0
3 5
𝟓𝟎𝟎
𝐅𝐂𝐃 = 𝐨𝐫 𝟏𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐓)
𝟑
 5 4
3 5 3
3

EXAMPLE 3 (SOLUTION cont’d) 4 4

5

AH = 0

▪ Apply method of joints. Consider Joint D. AV = 150 kN EV = 125 kN

Assume FDE and FDH as tension.
𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
Assume forces directed upward to be positive. 𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
σ FV = 0 𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
FDH
𝟐𝟓𝟎
−50 + FDH = 0 𝐅𝐅𝐂 = 𝐤𝐍 𝐓
𝟑
𝟖𝟎𝟎
𝐅𝐃𝐇 = 𝟓𝟎 𝐤𝐍 (𝐓) 𝐅𝐅𝐆 = 𝐤𝐍 𝐂
𝟑
500/3 FDE 𝐅𝐆𝐂 = 𝟎
Assume forces directed to the right to be positive. 𝟖𝟎𝟎
𝐅𝐆𝐇 = 𝐤𝐍 (𝐂)
𝟑
σ FH = 0 𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓)
𝟓𝟎𝟎
500 𝐅𝐂𝐃 = 𝐤𝐍 (𝐓)
− + FDE = 0 𝟑
3
𝟓𝟎𝟎
𝐅𝐃𝐄 = 𝐨𝐫 𝟏𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐓)
𝟑
 5 4
3 5 3
3

EXAMPLE 3 (SOLUTION cont’d) 4 4

5

AH = 0

▪ Apply method of joints. Consider Joint H. AV = 150 kN EV = 125 kN

Assume FHE as tension. Assume forces directed upward to be positive. 𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
σ FV = 0 𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
3 3 𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
−125 − 50 − FHE ( ) = 0 𝟐𝟓𝟎
5 5
𝐅𝐅𝐂 = 𝐤𝐍 𝐓
625 𝟑
800/3 FHE = − or − 208.3333 kN 𝟖𝟎𝟎
3
𝐅𝐅𝐆 = 𝐤𝐍 𝐂
𝟔𝟐𝟓 𝟑
FHE 𝐅𝐇𝐄 = 𝐨𝐫 𝟐𝟎𝟖. 𝟑𝟑𝟑𝟑 𝐤𝐍 (𝐂) 𝐅𝐆𝐂 = 𝟎
125 𝟑
𝟖𝟎𝟎
50 𝐅𝐆𝐇 = 𝐤𝐍 (𝐂)
Assume forces directed to the right to be positive. 𝟑
𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓)
σ FH = 0 𝟓𝟎𝟎
𝐅𝐂𝐃 = 𝐤𝐍 (𝐓)
𝟑
800 4 625 4 𝐅𝐃𝐇 = 𝟓𝟎 𝐤𝐍 (𝐓)
− 125 − =0 𝟓𝟎𝟎
3 5 3 5
𝐅𝐃𝐄 = 𝐤𝐍 (𝐓)
𝟑
0=0
 5 4
3 5 3
3

EXAMPLE 3 (SOLUTION cont’d) 4 4

5

AH = 0

▪ Apply method of joints. Consider Joint E. AV = 150 kN EV = 125 kN

𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
Assume forces directed upward to be positive.
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
σ FV = 0 𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
625/3 625 3 𝟐𝟓𝟎
− + 125 = 0 𝐅𝐅𝐂 = 𝐤𝐍 𝐓
3 5 𝟑
500/3 𝟖𝟎𝟎
0=0 𝐅𝐅𝐆 = 𝐤𝐍 𝐂
𝟑
𝐅𝐆𝐂 = 𝟎
𝟖𝟎𝟎
Assume forces directed to the right to be positive. 𝐅𝐆𝐇 = 𝐤𝐍 (𝐂)
125 𝟑
𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓)
σ FH = 0 𝟓𝟎𝟎
𝐅𝐂𝐃 = 𝐤𝐍 (𝐓)
625 4 500 𝟑
(
3 5
) − 3
=0 𝐅𝐃𝐇 = 𝟓𝟎 𝐤𝐍 (𝐓)
𝟓𝟎𝟎
𝐅𝐃𝐄 = 𝐤𝐍 𝐓
0=0 𝟑
𝟔𝟐𝟓
𝐅𝐇𝐄 = 𝐤𝐍 (𝐂)
𝟑
EXAMPLE 3 (SOLUTION cont’d)
𝐅𝐀𝐅 = 𝟐𝟓𝟎 𝐤𝐍 𝐂
𝐅𝐀𝐁 = 𝟐𝟎𝟎 𝐤𝐍 𝐓
𝐅𝐁𝐅 = 𝟏𝟎𝟎 𝐤𝐍 𝐓
𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓)
𝟐𝟓𝟎
𝐅𝐅𝐂 = 𝐤𝐍 𝐓
4 𝟑
𝟖𝟎𝟎
𝐅𝐅𝐆 = 𝐤𝐍 𝐂
𝟑
AH = 0 𝐅𝐆𝐂 = 𝟎
𝟖𝟎𝟎
𝐅𝐆𝐇 = 𝐤𝐍 (𝐂)
𝟑
𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓)
𝟓𝟎𝟎
AV = 150 kN EV = 125 kN 𝐅𝐂𝐃 = 𝐤𝐍 (𝐓)
𝟑
𝐅𝐃𝐇 = 𝟓𝟎 𝐤𝐍 (𝐓)
𝟓𝟎𝟎
𝐅𝐃𝐄 = 𝐤𝐍 𝐓
𝟑
𝟔𝟐𝟓
𝐅𝐇𝐄 = 𝐤𝐍 (𝐂)
𝟑
METHOD OF SECTIONS

Select a section that passes through as many members as possible whose

1 forces are desired, but not more than three members with unknown forces.

Draw the free-body diagram of the portion of the truss selected, showing all
external loads and reactions applied to it and the forces in the members that
2 have been cut by the section.

Determine the unknown forces by applying the three equilibrium equations:

3 σ FH = 0 , σ FV = 0 and σ M = 0 .

Apply an alternative equilibrium equation, which was not used to compute

4 member forces, to check the calculations.
EXAMPLE 4

Determine the force in members FG, FC, BC, GH, CH and CD of the Warren truss
shown below.
EXAMPLE 4 (SOLUTION)
▪ Draw the FBD. Assume the direction of the unknown reactions as shown below.

AH

AV EV

▪ Evaluate the static determinacy of the beam. r = 3; m = 13; j = 8

m + r ? 2j
13 + 3 ? 2(8)
16 = 16 (statically determinate)
EXAMPLE 4 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown
below.

AH

AV EV
Assume forces directed to the right to be positive.
σ FH = 0
𝐀𝐇 = 𝟎
EXAMPLE 4 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown below.

AH = 0

AV EV
Assume clockwise moments to be positive.
σ MA = 0
100 6 + 125 12 + 50 18 − EV 24 = 0
EV = 125 kN (upward)
 EXAMPLE 4 (SOLUTION cont’d)
▪ Considering the entire truss, apply the equations of equilibrium to calculate the
unknown reactions. Assume the direction of the unknown reactions as shown below.

AH = 0

AV EV = 125 kN
Assume forces directed upward to be positive.
σ FV = 0
AV − 100 − 125 − 50 + 125 = 0
AV = 150 kN (upward)
 EXAMPLE 4 (SOLUTION cont’d)
▪ Pass a section that cuts through members FG, FC and BC.
x

5 4
3 5
3 3
4 5
4

AH = 0

AV = 150 kN EV = 125 kN

x
 EXAMPLE 4 (SOLUTION cont’d)
▪ Consider the left portion of x-x and draw its FBD. Apply the three equations of equilibrium to
solve for the unknown member forces. Assume FFG, FFC and FBC as tension.
Assume forces directed upward to be positive.
FFG
3
5 σ FV = 0
4 FFC
3
FBC 150 − 100 − FFC =0
AH = 0 5
𝟐𝟓𝟎
𝐅𝐅𝐂 = 𝐨𝐫 𝟖𝟑. 𝟑𝟑𝟑𝟑 𝐤𝐍 (𝐓)
𝟑
AV = 150 kN
Assume clockwise moments to be positive. Assume forces directed to the right to be positive.
σ FH = 0
σ MF = 0
250 4
0 + FFG + + 200 = 0
150(6) − FBC (4.5) = 0 3 5
800
FFG = − or − 266.6667 kN
𝐅𝐁𝐂 = 𝟐𝟎𝟎 𝐤𝐍 (𝐓) 3
𝟖𝟎𝟎
𝐅𝐅𝐆 = 𝐨𝐫 𝟐𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐂)
𝟑
 EXAMPLE 4 (SOLUTION cont’d)
▪ Pass a section that cuts through members GH, CH and CD.
y

5 4
3 5
3 3
4 5
4

AH = 0

AV = 150 kN EV = 125 kN

y
 EXAMPLE 4 (SOLUTION cont’d)
▪ Consider the right portion of y-y and draw its FBD. Apply the three equations of equilibrium to
solve for the unknown member forces. Assume FGH, FCH and FCD as tension.
FGH Assume clockwise moments to be positive.
4
FCH 5
3 3 σ MC = 0
5
4
−FGH 4.5 + 50 6 − 125(12) = 0
FCD
800
FGH = − or − 266.6667 Kn
3
𝟖𝟎𝟎
EV = 125 kN 𝐅𝐆𝐇 = 𝐨𝐫 𝟐𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐂)
𝟑
Assume clockwise moments to be positive. Assume forces directed upward to be positive.
σ MH = 0 σ FV = 0
FCD 4.5 − 125(6) = 0 3
−FCH + 125 − 50 = 0
5
𝟓𝟎𝟎
𝐅𝐂𝐃 = 𝐨𝐫 𝟏𝟔𝟔. 𝟔𝟔𝟔𝟕 𝐤𝐍 (𝐓)
𝟑 𝐅𝐂𝐇 = 𝟏𝟐𝟓 𝐤𝐍 (𝐓)
EXAMPLE 5

Determine the forces in members CD, DG, and GH of the truss shown below by
the method of sections.
 EXAMPLE 4 (SOLUTION cont’d)
▪ Pass a section that cuts through members CD, DG and GH.
x

5
3
4

x
 EXAMPLE 5 (SOLUTION cont’d)
▪ Consider the right portion of x-x and draw its FBD. Apply the three equations of equilibrium to
solve for the unknown member forces. Assume FCD, FDG and FGH as tension.
Assume forces directed upward to be positive.
FGH σ FV = 0
FDG 4
3
FCD 5
3 FDG − 120 − 60 = 0
5

𝐅𝐃𝐆 = 𝟑𝟎𝟎 𝐤𝐍 (𝐓)

Assume clockwise moments to be positive. Assume forces directed to the right to be positive.

σ MD = 0 σ FH = 0
4
−FGH 3 + 60(4) = 0 −80 − 300 − FCD = 0
5

𝐅𝐆𝐇 = 𝟖𝟎 𝐤𝐍 (𝐓) FCD = −320 kN

𝐅𝐂𝐃 = 𝟑𝟐𝟎 𝐤𝐍 (𝐂)
REFERENCES

▪ Hibbeler, R.C. (2012). Structural Analysis (Eight Edition). Pearson Prentice

Hall.
▪ Kassimali, A. (2011). Structural Analysis (4th ed.) SI. Cengage Learning.