**Question**

A frame is loaded as shown above. There are internal hinges at nodes B and 2. Plot the internal stresses diagram due to the externally applied load.

**Solution**

Check for static determinacy.

A quick but non-general way of verifying degree of static indeterminacy of simple frames is;

D = R – e – S

where;

D = Degree of static indeterminacy

R = Number of support reactions = 5

e = Number of static equilibrium equations = 3

S = Number of special condition = 2 ( internal hinges)

D = 5 – 3 – 2 = 0

Therefore, the structure is statically determinate and stable.

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**Support Reactions**

**Let ∑F _{X} = 0**

Ax + 5 = 0

Ax = -5 KN ←

**Let ∑M _{2}^{R} = 0**

2Dy – (10 × 3) = 0

Dy = 15 KN

**Let ∑M _{B}^{R} = 0**

6Dy + 3Cy – (10 × 7) – ((2 × 4

^{2})/2) = 0

6(15) + 3Cy – 70 – 16 = 0

Cy = -1.333KN↓

**Let ∑M _{B}^{B} = 0**

Note that the due to the direction of the horizontal reaction A, it will exert a positive moment about column AB. Therefore;

4Ax – (5 × 2) – M

_{A}= 0

4(5) – 10 = M

_{A}

M

_{A}= 10 KN.m

**Let ∑M _{2}^{L} = 0**

4Ay + 4Ax + Cy – M

_{A}– ((2 × 4

^{2})/2) – (5 × 2) = 0

4Ay + 4(5) – 1.333 – 10 – 16 – 10 = 0

Ay = 4.333 KN

Equilibrium check;

∑F_{Y} ↓ = (2 × 8) + 10 = 18KN

∑F_{Y} ↑ = 15 – 1.333 + 4.333 = 18KN OK

Fig 1: Support Reactions Due to Externally Applied Load |

**Analysis of Internal forces**

**Section A – 1 (0 ≤ y ≤ 2.0m)**

(i) Bending Moment

My = 5y – 10 ———— (1)

*At y = 0*;

M

_{A}= -10 kNm

*At y = 2.0m;*

M

_{1}

^{B}= 5(2) – 10 = 0

(ii) Shear Force

The section is subjected to constant shear force;

Qx = ∂M_{y}/∂y = 5 KN

(iii) Axial Force

Ny + 4.333 = 0

N_{A-1} = -4.333 KN (compression)

**Section 1 – B (2.0m ≤ y ≤ 4.0m)**

(i) Bending Moment

M

_{y}= 5y – 5(y – 2) – 10 ———— (2)

*At y = 2m;*

M

_{1}

^{UP}= 5(2) – 10 = 0

*At y = 4.0m;*

M

_{B}

^{B}= 5(4) – 5(2) – 10 = 0

This shows that there is no bending moment at this section.

(ii) Shear Force

Differentiating equation (1) above yields zero. Therefore, there is shear force at the section also.

(iii) Axial Force

Ny + 4.333 = 0

N_{1-B} = -4.333 KN (compression)

**Section B – C (0 ≤ x ≤ 3.0m)**

(i) Bending Moment

M

_{x}= 4.333

*x*–

*x*

^{2}———— (3)

*At x = 0;*

M

_{B}

^{R}= 0

*At x = 3.0m;*

M

_{C}

^{L}= 4.333(3) – (3)2 = 4 KNm

Maximum moment occurs at the point of zero shear. Therefore;

Qx = ∂M_{x}/∂x = 4.333 – 2*x* = 0

On solving;

*x* = 4.333/2 = 2.1665m

M*max* = 4.333(2.1665) – (2.1665)2 = 4.693 KNm

(ii) Shear Force

The section is subjected to constant shear force;

Qx = ∂M_{x}/∂x = 4.333 – 2x

*At x = 0;*

Q_{B}_{R} = 4.333 KN

*At x = 3.0m;*

M_{C}^{L} = 4.333 – 2(3) = -1.667 KN

(iii) Axial Force

Nx = 0

No axial force

**Section C – 2 (3.0m ≤ x ≤ 4.0m)**

(i) Bending Moment

Mx = 4.333

*x*– 1.333(

*x*– 3) –

*x*

^{2}———— (4)

*At x = 3m;*

M

_{C}

^{R}= 4.333(3) – 32 = 4 KNm

*At x = 4.0m;*

M

_{2}= 4.333(4) – 1.333(1) – (4)2 = 0

(ii) Shear force

Expanding equation (4) above;

M_{x} = –*x*^{2} + 3*x* + 4

Qx = ∂Mx/∂x = -2x + 3

*At x = 3m*

Q_{C}^{R} = -2(3) + 3 = -3 KN

*At x = 4m*

Q_{2}^{L} = -2(4) + 3 = -5 KN

(iii) Axial Force

No axial force

**Section D – E (0 ≤ x ≤ 1.0)**

(i) Bending Moment

M

_{x}= -10

*x*————— (5)

*At x = 0;*

M_{E} = 0

*At x = 1.0m;*

M_{C}^{L} = -10(1) = -10 KNm

(ii) Shear Force

Qx = ∂M_{x}/∂x = -10 KN

But since we are coming from the right to the left, downward force is positive.

Therefore the section subjected to a positive constant shear force of 10 KN

(iii) Axial force

There is no axial force at the section

**Internal Stresses Diagram**

We can now plot the internal stresses diagram from the values obtained above.

Fig 2: Final Bending Moment Diagram |

** **

Fig 3: Final Shear Force Diagram |

Fig 4: Final Axial Force Diagram |

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