University of Duhok

College of Engineering

Water Resources Engineering

Assignment 4
Analysis of Compound Trusses

Student Name: Ammar Mohammad Abdullah

Stage: 3
Course: Theory of structural
Instructor: Gulan Bapeer Hassan
 Problem
Find all member forces of the compound truss shown in figure below
a

Solution:

∑ 𝑀𝐼 = 0 ;
+

−(130 × 3) − (130 × 6) − (130 × 9) − (130 × 12) − (130 × 15) − (130 × 18) − (130 × 21)
+𝐴𝑦(24) + (90 × 3) + (45 × 6) = 0
𝐴𝑦 = 432.5kN

+ ∑ 𝐹𝑥 = 0
45 + 90 − 𝐴𝑥 = 0
𝐴𝑥 = 135 𝐾𝑁

∑ 𝐹𝑦 = 0
+

−7(130) + 432.5 + 𝐼𝑦 = 0
𝐼𝑦 = 477.5 𝐾𝑁

• We can obtain the force in NO by using section aa in Fig.Below

∑ 𝑀𝐸 = 0 ;
+

−(130 × 3) − (130 × 6) − (130 × 9) + (432.5 × 12) + (90 × 3) + (45 × 6) − 𝐹𝑁𝑂 (6) = 0

𝐹𝑁𝑂 = 565 𝐾𝑁 (Compression)

a 𝐹𝑁𝑂

3m

3m

E
a

𝐴𝑦 = 432.5kN
 • Joint A y
FAJ
∑ 𝐹𝑦 = 0;
+

432.5 − 𝐹𝐴𝐽 sin 45 = 0 45°

135 KN FAB
x
𝐹𝐴𝐽 = 611.69 𝐾𝑁 (Compression) (A)

+ ∑ 𝐹𝑥 = 0 ;
432.5 KN
−135 + 𝐹𝐴𝐵 − 611.69 cos 45 = 0
𝐹𝐴𝐵 = 567.5 𝐾𝑁 (Tension)

• Joint B
∑ 𝐹𝑦 = 0;
+

FBJ
−130 − 𝐹𝐵𝐽 = 0 y

𝐹𝐵𝐽 = 130𝐾𝑁 (Tension) 567.5 KN (B)

FBC
x
+ ∑ 𝐹𝑥 = 0 ;
− 567.5 + 𝐹𝐵𝐶 = 0
130 KN
𝐹𝐵𝐶 = 567.5 𝐾𝑁 (Tension)

• Joint J x
y FJN
∑ 𝐹𝑦 = 0;
+

𝐹𝐽𝐶 − 130 cos 45 − 90 sin 45 = 0 90 KN

(J)
𝐹𝐽𝐶 = 155.6 𝐾𝑁 (Compression) 45°
45°
FJC
611.69 KN
+ ∑ 𝐹𝑥 = 0 ; 130 KN

− 𝐹𝐽𝑁 + 611.69 − 130 sin 45 + 90 cos 45 = 0

𝐹𝐽𝑁 = 583.41 𝐾𝑁 (Compression)
 y

• Joint N
+ ∑ 𝐹𝑥 = 0 ; 45 KN (N) 565 KN
x
−565 + 𝐹𝑁𝐾 cos 45 + 45 + 583.4 sin 45 = 0 45°

𝐹𝑁𝐾 = 152 (Tension) 45° FNK

FNC
∑ 𝐹𝑦 = 0; 583.4KN
+

583.4 cos 45 − 𝐹𝑁𝐶 − 152 sin 45 = 0

𝐹𝑁𝐶 = 305 𝐾𝑁 (Tension)
305 KN
y FCK
155.6 KN
• Joint C
+ ∑ 𝐹𝑥 = 0 ; 45°
45°

−130 − 𝐹𝐶𝐾 sin 45 + 305 − 155.6 sin 45 = 0 567.5KN x (C) FCD

𝐹𝐶𝐾 = 91.89(Compression)
∑ 𝐹𝑦 = 0; 130 KN
+

−567.5 + 𝐹𝐶𝐷 + 155.6 cos 45 − 91.89 cos 45 = 0

𝐹𝐶𝐷 = 522.5 𝐾𝑁 (Tension)

• Joint D
+ ∑ 𝐹𝑥 = 0 ; FDK
y
𝐹𝐷𝐸 = 522.5(Tension)
522.5 KN (D)
FDE
x
∑ 𝐹𝑦 = 0;
+

𝐹𝐷𝐾 = 130 𝐾𝑁 (Tension)

130 KN
 • Joint K
152 KN y
+ ∑ 𝐹𝑥 = 0 ;
−152 + 𝐹𝐾𝐸 + 130 cos 45 = 0 (K)

𝐹𝐾𝐸 = 60.1 𝐾𝑁(Tension)

FKE
45°
901.89KN
x
130 KN
• Joint O
∑ 𝐹𝑦 = 0;
+

𝐹𝐶𝐸 = 0 y

565 KN (O)
+ ∑ 𝐹𝑥 = 0 ; FOP
x
𝐹0𝑃 = 565 𝐾𝑁 (Compression)
FOE

y FEL
• Joint E 60.1 KN 0

∑ 𝐹𝑦 = 0; 45° 45°
+

522.5 𝐾𝑁
FEF
−130 + 60.1 sin 45 + 𝐹𝐸𝐿 sin 45 = 0 x (E)

𝐹𝐸𝐿 = 123.75 𝐾𝑁 (Tension)

130 KN
+ ∑ 𝐹𝑥 = 0 ;
−522.5 + 123.75 cos 45 + 𝐹𝐸𝐹 − 60.1 cos 45 = 0
𝐹𝐸𝐹 = 477.5(Tension)
 • Joint F
+ ∑ 𝐹𝑥 = 0 ; FFL
y
𝐹𝐹𝐺 = 477.5 (Tension)
477.5 KN (F)
FFG
x
∑ 𝐹𝑦 = 0;
+

𝐹𝐹𝐿 = 130 𝐾𝑁 (Tension)

130 KN

• Joint L x
y FLP
+ ∑ 𝐹𝑥 = 0 ;
−123.75 + 𝐹𝐿𝑃 − 130 sin 45 = 0
(L)
𝐹𝐿𝑃 = 215.7 𝐾𝑁(Tension)
FLG
∑ 𝐹𝑦 = 0;
+

45°
123.75 KN
−130 cos 45 + 𝐹𝐿𝐺 = 0 130 KN
𝐹𝐿𝐺 = 91.9 𝐾𝑁 (Compression)

• Joint P
y
+ ∑
𝐹𝑋 = 0;
−𝐹𝑃𝑀 sin 45 + 565 − 215.7 sin 45 = 0
565 KN (P)
𝐹𝑃𝑀 = 583.4 𝐾𝑁 (Compression) x

∑ 𝐹𝑦 = 0 ;
+

45° 45°
FPM
−𝐹𝑃𝐺 + 583.4 cos 45 − 215.7 cos 45 = 0
215.7 KN FPG
𝐹𝑃𝐺 = 260 𝐾𝑁 (Tension)
 • Joint G
260 KN
y FGM
∑ 𝐹𝑦 = 0; 91.9 KN
+

−130 + 260 − 91.9 sin 45 − 𝐹𝐺𝑀 sin 45 = 0 45°

45°
477.5 𝐾𝑁
𝐹𝐺𝑀 = 91.9 𝐾𝑁 (Compression) FGH
x (G)

+ ∑ 𝐹𝑥 = 0 ;
−477.5 + 91.9 cos 45 + 𝐹𝐺𝐻 − 91.9 cos 45 = 0
130 KN
𝐹𝐺𝐻 = 477.5(Tension)

• //Joint H
+ ∑ 𝐹𝑥 = 0 ; FHM
y
𝐹𝐻𝐼 = 477.5 (Tension)
477.5 KN
∑ 𝐹𝑦 = 0; (H)
+

FHI
x
𝐹𝐻𝑀 = 130 𝐾𝑁 (Tension)

130 KN
• Joint I FIM y

∑ 𝐹𝑦 = 0;
+

477.5 − 𝐹𝐼𝑀 sin 45 = 0 45°

x
𝐹𝐼𝑀 = 675.3 𝐾𝑁 (Compression) 477.5 KN (I)

477.5 KN