Structural Analysis of Determinate Arch-Frame Compound Structure

∑M_F^L = 0
8Dy – 4Dx – (10 × 8²)/2 = 0
8(80) – 4Dx – (10 × 8²)/2 = 0
– 4Dx + 320 = 0
Therefore, Dx = 320/4 = 80 KN = Ex

Geometric Properties of the Arch Section

The ordinate of the arch at any given horizontal length section is given by;

y = [4y_c (Lx – x²)] / L²

Where y_c is the height of the crown of the arch
y = [(4 × 4) × (16x – x²)]/16² = x – (x²/16)

dy/dx = y’ = 1 – x/8

At x = 0; y = 0
y’ = 1 – 0/8 = 1
sin⁡θ = y’/√(1 + y’²) = 1/√(1 + 1²) = 0.7071
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 1²) = 0.7071

At x = 4m;
y = x – x²/16 = 4 – (4²/16) = 3m
y’ = 1 – 4/8 = 0.5
sin⁡θ = y’/√(1 + y’²) = 0.5/√(1 + 0.5²) = 0.4472
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0.5²) = 0.8944

At x = 8m;
y = x – x²/16 = 8 – (8²/16) = 4m
y’ = 1 – 8/8 = 0
sin⁡θ = y’/√(1 + y’²) = 0/√(1 + 0.5² ) = 0
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0) = 1.0

At x = 12m;
y = x – x²/16 = 12 – (12²/16) = 3m
y’= 1 – x/8 = 1 – 12/8 = -0.5
sin⁡θ = y’/√(1 + y’²) = (-0.5)/√(1 + 0.5²) = -0.4472
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 0.5² ) = 0.8944

At x = 16m;
y = x – x²/16 = 16 – (16²/16) = 0
y’= 1 – 16/8 = 1 – 12/8 = -1.0
sin⁡θ = y’/√(1 + y’²) = (-1.0)/√(1 + 1² ) = -0.7071
cos⁡θ = 1/√(1 + y’²) = 1/√(1 + 1² ) = 0.7071

Internal Stresses in the Arch StructureBending Moment
M_D = 0
M₁ = (80 × 4) – (80 × 3) – (10 × 4²)/2 = 0
M_F = (80 × 8) – (80 × 4) – (10 × 8²)/2 = 0
M₂ = (80 × 12) – (80 × 3) – (10 × 12²)/2 = 0
M_E = 0

Shear force
Q_i = ∑V cos⁡θ – ∑H sin⁡θ
Q_D = (80 ×0.7071) – (80 × 0.7071) = 0 (No shear)
Q₁^L = Q₁^R [80 – (10 × 4)] × 0.8944 – (80 × 0.4472) = 0 (No shear)
Q_F = [80 – (10 × 8)] × 1.0 – (80 × 0) = 0 (No shear)
Q₂^L = Q₃^R = [80 – (10 × 12)] × 0.8944 – (80 × -0.4472) = 0 (No shear)
Q_E = [80 – (10 × 16)] × 0.7071 – (80 × -0.7071) = 0 (No shear)

Axial force
N_i = -∑V sin⁡θ – ∑H cos⁡θ
N_D = -(80 × 0.7071) – (80 × 0.7071) = -113.136 KN (Compression)
N₁^L = -[80 – (10 × 4)] × 0.4472 – (80 × 0.8944) = -89.44 KN
N_F = -[80 – (10 × 8)] × 0 – (80 × 1) = -80 KN (Compression)
N₂^L = N₃^R = -[80 – (10 × 12)] × -0.4472 – (80 × 0.8944) = -89.44 KN
N_E = -[80 – (10 × 16)] × -0.7071 – (80 × 0.7071) = -113.136 KN

We now transfer the reactive forces from the arch section to the the supporting trusses below;

θ = tan^-1⁡(6/8) = 36.869°
cos⁡θ = 0.8
sin⁡θ = 0.6

Analysis of Joint D

∑F_X = 0
-35 – 0.8F_AD + 0.8F_DB = 0
– 0.8F_AD + 0.8F_DB = 35 —————- (1)

∑F_Y = 0
-80 – 0.6F_AD – 0.6F_DB = 0
– 0.6F_AD – 0.6F_DB = 80 —————- (2)

Solving (1) and (2) simultaneously;
F_AD = -88.542 KN (Compression)
F_DB = -44.792 KN (Compression)

Analysis of joint E

∑F_X = 0
80 – 0.8F_BE + 0.8F_EC = 0
– 0.8F_BE + 0.8F_EC = 80 —————- (3)

∑F_Y = 0
-80 – 0.6F_BE – 0.6F_EC = 0
– 0.6F_BE – 0.6F_EC = 80 —————- (4)

Solving (3) and (4) simultaneously;
F_BE = -16.667 KN (Compression)
F_DB = -116.667 KN (Compression)

Support Reactions

Support A

∑F_Y = 0
Ay + 0.6F_AD = 0
Ay – (0.6 × 88.542) = 0
Ay = 53.1252 KN

∑F_X = 0
Ax + 0.8F_AD = 0
Ax – (0.8 × 88.542) = 0
Ax = 70.8336 KN

Support B

∑F_X = 0
Bx – 0.8F_BD + 0.8F_BE = 0
Bx – (0.8 × -44.792) + (0.8 × -16.667) = 0
Bx = -22.5 KN

∑F_Y = 0
By + 0.6F_BD + 0.6F_BE = 0
By – (0.6 × 44.792) – (0.6 × 16.667) = 0
By = -36.8754 KN

Support C

∑F_Y = 0
Cy + 0.6F_EC = 0
Cy – (0.6 × 116.667) = 0
Cy = 70 KN

∑F_X = 0
-Cx – 0.8F_EC = 0
-Cx – (0.8 × -116.667) = 0
Cx = 93.333 KN

Equilibrium Verification

∑F_Y ↓ = (10 × 16) = 160 KN
∑F_Y ↑ = Ay + By + Cy = 53.1252 + 36.8754 + 70 = 160 KN

∑F_X → = 45 + 70.833 = 115.833 KN
∑F_X ← = 22.5 + 93.333 = 115.833 KN

Internal Stresses Diagram

As you can see, there are no bending moment and shear forces on the structure, therefore, the axial force diagram is as given below;