**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 }+ 2.5^{2}) = 6.5m

**Notations:**

**Example: N _{C}^{R }= Axial Load at point C, just to the Right**

**Let ∑M _{G} = 0**

12Ay – [(10 × 12

^{2})/2] – (25 × 11) + (7 × 4) – (16 × 1) = 0

12Ay = 983

Ay = 81.917 kN

**Let ∑M _{D}^{L} = 0**

6Ay – 8Ax – (25 × 5) – (7 × 4) – [(10cos22.619° × 6.5

^{2})/2] = 0

6(81.917) – 8Ax – (25 × 5) – (7 × 4) – [(9.23 × 6.5

^{2})/2] = 0

8Ax = 143.518 kN

Ax = 17.939 kN

**Let ∑M _{A} = 0**

12Gy – [(10 × 12

^{2})/2] – (25 × 1) – (7 × 4) – (16 × 11) = 0

12Gy = 949

Gy = 79.083 kN

**Let ∑M _{D}^{R} = 0**

6Gy – 8Gx – (16 × 5) – [(10cos22.619° × 6.5

^{2})/2] = 0

6(79.083) – 8Gx – (16 × 5) – [(9.23 × 6.5

^{2})/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.

Read Also….

**Understanding Sign Conventions in Structural Analysis**

*(b) Internal Stresses*

*Section A – B ^{B} (0 ≤ y ≤ 4.0m)*

(i) Bending moment

M

*y*= -Ax.

*y*= -17.939

*y*

At

*y*= 0; M

_{A}= 0

At

*y*= 4.0m; M

_{B}

^{B}= (-17.939 × 4) = -71.756 kNm

(ii) Shear Force

Qy = ∂M*y*/∂y = -17.939 kN

(iii) Axial Force

Ny + Ay = 0

Ny = -Ay = -81.917 kN (compression)

*Section 1 – B ^{R} (0 ≤ x ≤ 1.0m)*

(i) Bending moment

M

*x*= -25.

*x*

At

*x*= 0; M

_{1}= 0

At

*x*= 1.0m; M

_{B}

^{R}= (-25 × 1) = -25 kNm

(ii) Shear force

Q*x* = ∂M*x*/∂x = -25 kN

(iii) Axial Force

N*x *– 7kN = 0

N*x* = 7kN (tension)

*Section **B ^{UP}*

*– C*

^{B}(4 ≤ y ≤ 5.5m)(i) Bending moment

M

*y*= -Ax.

*y +*(25 × 1) – 7(

*y*– 4)

At

*y*= 4m; M

_{B}

^{UP}= -(17.939 × 4) + 25 = -46.756 kNm

At

*y*= 5.5m; M

_{C}

^{B}= -(17.939 × 5.5) + 25 – 7(1.5) = -84.1645 kNm

Q

*x*= -17.939 – 7 = -24.939 kN

(iii) Axial force

Ny + Ay – 25kN = 0

N*y* = -81.917 + 25 = -56.917 kN (compression)

*Section*

*C*

^{R}*– D*

^{L}(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

M*z* = * *(∑V.cos22.619-9°.*z*) – (∑H.sin22.619°.*z*) – [(10cos22.619° ×* z*^{2})/2] – 84.1645

M*z* = * *52.539*z – *9.591*z – *4.6154z^{2} – 84.1645

M*z* = -4.6154*z*^{2 }*+ *42.9425*z *– 84.1645

At *z* = 0; M_{C}^{R} = -84.1645 kNm

At *z* = 6.5m; M_{C}^{B} = -4.6154*(6.5)*^{2 }*+ *42.948*(6.5) *– 84.1645 = -195 + 279.126 – 84.1645 = 0

*Maximum span moment*

M*z* = -4.6154*z*^{2 }*+ *42.9425*z *– 84.1645

Maximum moment occurs at the point of zero shear

∂M*z*/∂z = -9.23z + 42.9425 = 0

z = 42.9425/9.23 = 4.652m

Mmax = -4.6154*(4.652)*^{2 }*+ *42.9425*(4.652) *– 84.1645 = -99.882 + 199.768 – 84.1645 = 15.7215 kNm

*(ii) Shear force*

∂M*z*/∂z = (∑V.cos22.619) – (∑H.sin22.619)

At *z* = 0; Q_{C}^{R} = (56.917 × 0.923) – (24.939 × 0.385) = 42.933 kN

At *z* = 6.5m; Q_{D}^{L} = [(56.917 – (10 × 6)) × 0.923] – (24.939 × 0.385) = -2.8456 – 9.6015 = -12.447 kN

*(iii) Axial force*

*z*= -(∑V.sin22.619) – (∑H.cos22.619)

At *z* = 0; N_{C}^{R} = -(56.917 × 0.385) – (24.939 × 0.923) = -21.913 – 23.018 = -44.931 kN

At *z* = 6.5m; N_{D}^{L} = [-(56.917 – (10 × 6)) × 0.385] – (24.939 × 0.923) = 1.187 – 23.018 = -21.831 kN

**Coming from the right**

*Section G – F ^{B} (0 ≤ y ≤ 4.0m)*

(i) Bending moment

M

*y*= -Gx.

*y*= -24.939

*y*

At

*y*= 0; M

_{G}= 0

At

*y*= 4.0m; M

_{F}

^{B}= (-24.939 × 4) = -99.756 kNm

(ii) Shear Force

Qy = ∂M*y*/∂y = +24.939 kN (note that we are coming from right to left)

(iii) Axial Force

Ny + Gy = 0

*Section 2 – F ^{R} (0 ≤ x ≤ 1.0m)*

(i) Bending moment

M

*x*= -16.

*x*

At

*x*= 0; M

_{2}= 0

At

*x*= 1.0m; M

_{F}

^{R}= (-16 × 1) = -16 kNm

(ii) Shear force

Q*x* = ∂M*x*/∂x = -16 kN

(iii) Axial Force

N*x *– 0 = 0

*x*= 0 (no force)

*Section **F ^{UP}*

*– E*

^{B}(4 ≤ y ≤ 5.5m)(i) Bending moment

M

*y*= -Gx.

*y +*(16 × 1)

At

*y*= 4m; M

_{F}

^{UP}= -(24.939 × 4) + 16 = -83.756 kNm

At

*y*= 5.5m; M

_{E}

^{B}= -(24.939 × 5.5) + 16 = -121.1645 kNm

Q

*x*= 24.939 kN

(iii) Axial force

Ny + Ay – 16kN = 0

N*y* = -79.083 + 16 = -63.083 kN (compression)

*Section*

*E*

^{UP}*– D*

^{R}(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

M*z* = * *(∑V.cos22.619-9°.*z*) – (∑H.sin22.619°.*z*) – [(10cos22.619° ×* z*^{2})/2] – 121.1645

M*z* = * *58.225*z – *9.5915*z – *4.6154z^{2} – 121.1645

M*z* = -4.6154*z*^{2 }*+ *48.6335*z *– 121.1645

At *z* = 0; M_{E}^{R} = -121.1645 kNm

At *z* = 6.5m; M_{C}^{B} = -4.6154*(6.5)*^{2 }*+ *48.6335*(6.5) *– 121.1645 = -195 + 316.1175 – 121.1645 = 0

*Maximum span moment*

M*z* = -4.6154*z*^{2 }*+ *48.6335*z *– 121.1645

Maximum moment occurs at the point of zero shear

∂M*z*/∂z = -9.23z + 48.6335 = 0

z = 48.6335/9.23 = 5.269m

Mmax = -4.6154*(5.269)*^{2 }*+ *48.6335*(5.269) *– 121.1645 = -128.134 + 256.2499 – 121.1645 = 6.9514 kNm

*(ii) Shear force*

∂M*z*/∂z = -(∑V.cos22.619) + (∑H.sin22.619)

At *z* = 0; Q_{C}^{R} = (63.083 × 0.923) – (24.939 × 0.385) = -58.2256 + 9.6015 = -48.624 kN

At *z* = 6.5m; Q_{D}^{L} = -[(63.083 – (10 × 6)) × 0.923] + (24.939 × 0.385) = -2.8456 + 9.6015 = 6.756kN

*(iii) Axial force*

*z*= -(∑V.sin22.619) – (∑H.cos22.619)

At *z* = 0; N_{C}^{R} = -(63.083 × 0.385) – (24.939 × 0.923) = -24.287 – 23.018 = -47.305kN

At *z* = 6.5m; N_{D}^{L} = [-(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*

Its a very informative article for structural engineers.

You can get free structural engineering resources from my website:

https://worldcentre.me

Thanks

Mahmood

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