# Analysis of Industrial Gable Frames 1.0 Introduction
Industrialization is one of the major keys to development and sustainable economy. Industrial structures are usually very easy to identify because of their unique  features that are quite different from residential or commercial buildings. Engineers are often tasked with analysing and designing industrial frames, and to simplify the analysis for manual calculations, they are usually idealised as 2D plane frames.

2.0 Solved Example
In this post, a gable industrial frame structure is subjected to a load regime as shown below. The frame is hinged at point D, and it is desired to obtain the internal forces (bending moment, axial forces, and shear forces) due to the externally applied load.

N/B: In practical construction, it is not advisable to hinge the structure at the apex due to the problem of excessive deflection and stability of the frame.

### Solution

(a) Support Reactions

Geometrical Properties
Angle of inclination of the rafter
⁡θ = tan-1(2.5/6) = 22.619°
cos ⁡θ = 0.923
sin ⁡θ = 0.385

Length of rafter (z) = sqrt(6+ 2.52) = 6.5m

Notations:
Example: NC= Axial Load at point C, just to the Right

Let ∑MG = 0
12Ay – [(10 × 122)/2] – (25 × 11) + (7 × 4) – (16 × 1) = 0
12Ay = 983
Ay = 81.917 kN

Let ∑MDL = 0
6Ay – 8Ax –  (25 × 5) – (7 × 4) – [(10cos22.619° × 6.52)/2]  = 0
6(81.917) – 8Ax –  (25 × 5) – (7 × 4) – [(9.23 × 6.52)/2]  = 0
8Ax = 143.518 kN
Ax = 17.939 kN

Let ∑MA = 0
12Gy – [(10 × 122)/2] – (25 × 1) – (7 × 4) – (16 × 11) = 0
12Gy = 949
Gy = 79.083 kN

Let ∑MDR = 0
6Gy – 8Gx –  (16 × 5) –  [(10cos22.619° × 6.52)/2]  = 0
6(79.083) – 8Gx –  (16 × 5) – [(9.23 × 6.52)/2]  = 0
8Gx = 199.515 kN
Gx = 24.939 kN

Equilibrium Check
All downward vertical forces = (10 × 12) + 25 + 16 = 161 kN
Upward reactive forces = 81.917 + 79.083 = 161 kN

All rightward horizontal forces = Ax + 7kN = 17.939 + 7 = 24.939 kN
All leftward horizontal forces = Gx = 24.939 kN

Therefore, equilibrium is ok. If you are having confusions, the post below might help.

(b) Internal Stresses
Section A – BB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Ax.y = -17.939y
At y = 0; MA = 0
At y = 4.0m; MBB = (-17.939 × 4) = -71.756 kNm

(ii) Shear Force
Qy = ∂My/∂y = -17.939 kN

(iii) Axial Force
Ny + Ay = 0
Ny = -Ay = -81.917 kN (compression)

Section 1 – BR (0 ≤ x ≤ 1.0m)
(i) Bending moment
Mx = -25.x
At x = 0; M1 = 0
At x = 1.0m; MBR = (-25 × 1) = -25 kNm

(ii) Shear force
Qx = ∂Mx/∂x = -25 kN

(iii) Axial Force
N– 7kN = 0
Nx = 7kN  (tension)

Section BUP – CB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Ax.y + (25 × 1) – 7(y – 4)
At y = 4m; MBUP = -(17.939 × 4) + 25 = -46.756 kNm
At y = 5.5m; MCB = -(17.939 × 5.5) + 25 – 7(1.5) = -84.1645 kNm

(ii) Shear force
Qx + Ax + 7kN = 0
Qx = -17.939 – 7 =  -24.939 kN

(iii) Axial force
Ny + Ay – 25kN = 0
Ny = -81.917 + 25 =  -56.917 kN (compression)

Section CR – DL (0 ≤  z ≤ 6.5m) (Pay careful attention here)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -84.1645 kNm
Summation of vertical force transferred ∑V = Ay – 25 = 81.917 – 25 = 56.917 kN
Summation of horizontal force transferred ∑H = Ax + 7 = 17.939  + 7 = 24.939 kN

Mz =  (∑V.cos22.619-9°.z) – (∑H.sin22.619°.z) – [(10cos22.619° × z2)/2] – 84.1645
Mz =  52.539z – 9.591z – 4.6154z2 – 84.1645
Mz = -4.6154z42.9425– 84.1645

At z = 0; MCR = -84.1645 kNm
At z = 6.5m; MCB = -4.6154(6.5)42.948(6.5) – 84.1645 = -195 + 279.126 – 84.1645 = 0

Maximum span moment
Mz = -4.6154z42.9425– 84.1645

Maximum moment occurs at the point of zero shear
∂Mz/∂z = -9.23z + 42.9425 = 0
z = 42.9425/9.23 = 4.652m

Mmax  = -4.6154(4.652)42.9425(4.652) – 84.1645 = -99.882 + 199.768 – 84.1645 = 15.7215 kNm

(ii) Shear force
∂Mz/∂z = (∑V.cos22.619) – (∑H.sin22.619)

At z = 0; QCR = (56.917 × 0.923) – (24.939 × 0.385) = 42.933 kN
At z = 6.5m; QDL = [(56.917 – (10 × 6))   × 0.923] – (24.939 × 0.385) = -2.8456 – 9.6015 = -12.447 kN

(iii) Axial force

Nz = -(∑V.sin22.619) – (∑H.cos22.619)

At z = 0; NCR = -(56.917 × 0.385) – (24.939 × 0.923) = -21.913 – 23.018 =   -44.931 kN
At z = 6.5m; NDL = [-(56.917 – (10 × 6))  × 0.385] – (24.939 × 0.923) = 1.187 – 23.018 = -21.831  kN

Coming from the right
Section G – FB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Gx.y = -24.939y
At y = 0; MG = 0
At y = 4.0m; MFB = (-24.939 × 4) = -99.756 kNm

(ii) Shear Force
Qy = ∂My/∂y = +24.939 kN (note that we are coming from right to left)

(iii) Axial Force
Ny + Gy = 0

Ny = -Gy = -79.083 kN (compression)

Section 2 – FR (0 ≤  x ≤ 1.0m)
(i) Bending moment
Mx = -16.x
At x = 0; M2 = 0
At x = 1.0m; MFR = (-16 × 1) = -16 kNm

(ii) Shear force
Qx = ∂Mx/∂x = -16 kN

(iii) Axial Force
N– 0 = 0

Nx = 0 (no force)

Section FUP – EB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Gx.y + (16 × 1)
At y = 4m; MFUP = -(24.939 × 4) + 16 = -83.756 kNm
At y = 5.5m; MEB = -(24.939 × 5.5) + 16 = -121.1645 kNm

(ii) Shear force
Qx – Gx = 0
Qx =  24.939 kN

(iii) Axial force
Ny + Ay – 16kN = 0
Ny = -79.083 + 16 =  -63.083 kN (compression)

Section EUP – DR (0 ≤  z ≤ 6.5m)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -121.1645 kNm
Summation of vertical force transferred ∑V = Gy – 16 = 79.083 – 16 = 63.083 kN
Summation of horizontal force transferred ∑H = Gx = 24.939 kN

Mz =  (∑V.cos22.619-9°.z) – (∑H.sin22.619°.z) – [(10cos22.619° × z2)/2] – 121.1645
Mz =  58.225z – 9.5915z – 4.6154z2 – 121.1645
Mz = -4.6154z48.6335– 121.1645

At z = 0; MER = -121.1645 kNm
At z = 6.5m; MCB = -4.6154(6.5)48.6335(6.5) – 121.1645 = -195 + 316.1175 – 121.1645 = 0

Maximum span moment
Mz = -4.6154z48.6335– 121.1645

Maximum moment occurs at the point of zero shear
∂Mz/∂z = -9.23z + 48.6335 = 0
z = 48.6335/9.23 = 5.269m

Mmax  = -4.6154(5.269)48.6335(5.269) – 121.1645 = -128.134 + 256.2499 – 121.1645 = 6.9514 kNm

(ii) Shear force
∂Mz/∂z = -(∑V.cos22.619) + (∑H.sin22.619)

At z = 0; QCR = (63.083 × 0.923) – (24.939 × 0.385) = -58.2256 + 9.6015 =  -48.624 kN
At z = 6.5m; QDL = -[(63.083 – (10 × 6)) × 0.923] + (24.939 × 0.385) = -2.8456 + 9.6015 = 6.756kN

(iii) Axial force

Nz = -(∑V.sin22.619) – (∑H.cos22.619)

At z = 0; NCR = -(63.083 × 0.385) – (24.939 × 0.923) = -24.287 – 23.018 =  -47.305kN
At z = 6.5m; NDL = [-(63.083 – (10 × 6))  × 0.385] – (24.939 × 0.923) = -1.187 – 23.018 = -24.205 kN

(b) Internal Stresses Diagram
(a) Bending Moment Diagram

(b) Shear Force Diagram

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7. Gerald Ezeobi

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I used both Etabs for the analysis and Protastructure. I compared results to detail the final output.
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