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BAC2 tech · Technology and Engineering II
Mechanical systems · Structures

Calculation of beams and trusses

Knowing the reactions of a beam is only the first step. To size it, you need to know how the internal forces are distributed along its whole length, and in exactly which section they are greatest: that is where the member needs more cross-section or more material.

Before you start, make sure you know: Forces, loads and stresses (2nd ESO)
Before you start, make sure you know: Solving right-angled triangles (1st Bachillerato)

Shear force and bending moment diagrams

Simply supported beam with a central point load and its corresponding diagrams: stepped constant shear force and triangular bending moment V shear M moment
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Explain what internal force diagrams of a beam are and what they are used for: what the shear force diagram represents and what the bending moment diagram represents,…

Explain what internal force diagrams of a beam are and what they are used for: what the shear force diagram represents and what the bending moment diagram represents, and why the cross-section sized is the one where the bending moment is maximum (the "critical section").

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Distinguish between shear force and bending moment in a beam cross-section: what each tends to do (sliding one part vertically relative to the other versus bending the…

Distinguish between shear force and bending moment in a beam cross-section: what each tends to do (sliding one part vertically relative to the other versus bending the member) and what SI units each is measured in.

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Explain the common sign convention for bending moments: positive when the beam bends with the bottom face in tension and negative when it bends the other way. Explain on…

Explain the common sign convention for bending moments: positive when the beam bends with the bottom face in tension and negative when it bends the other way. Explain on which face of a simply supported reinforced concrete beam the main reinforcement bars are placed and why.

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Simply supported beam with a point load

Simply supported beam with an off-centre point load P, with the shear force diagram V constant in stretches (+1 200 N and −800 N) and the bending moment diagram M triangular with its maximum under the load P A B a b V +1 200 N −800 N M M max = P·a·b/L 0 0
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The figure shows a simply supported beam of $5\ \text{m}$ with a point load of $2\,000\ \text{N}$ applied at $2\ \text{m}$ from support A. a) Calculate the reactions at…

The figure shows a simply supported beam of $5\ \text{m}$ with a point load of $2\,000\ \text{N}$ applied at $2\ \text{m}$ from support A. a) Calculate the reactions at A and B. b) Draw by hand the shear force diagram indicating its value in each section. c) State the value of the maximum bending moment and the section at which it occurs.

A B P = 2000 N RA = ? RB = ? 2 m 3 m

You can do it on paper and send a photo with your answer.

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A simply supported beam of $6\ \text{m}$ carries a load of $3\,000\ \text{N}$ at its centre. a) Calculate the reactions. b) Draw the bending moment diagram (a…

A simply supported beam of $6\ \text{m}$ carries a load of $3\,000\ \text{N}$ at its centre. a) Calculate the reactions. b) Draw the bending moment diagram (a symmetrical triangle) and calculate the maximum moment using $M_{max} = \frac{P \cdot L}{4}$. c) Explain why the shear force diagram changes sign right at the section where the moment is maximum.

You can do it on paper and send a photo with your answer.

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Based on the beam and the internal force diagram schematics shown in the figure: a) Determine the reactions at supports A and B using the fundamental equilibrium…

Based on the beam and the internal force diagram schematics shown in the figure:

  1. a) Determine the reactions at supports A and B using the fundamental equilibrium equations of statics and deduce the exact numerical values missing from the shear force diagram.
  2. b) Analytically prove the value of the maximum bending moment in the section under the load without using memorized formulas, by deducing and checking the relationship between the area of the shear force diagram and the change in bending moment.
  3. c) If the beam is made of steel with an allowable yield strength $\sigma_{\text{adm}} = 150\ \text{MPa}$, calculate the minimum section modulus ($W_z$) required to ensure safety against bending failure. Reasonably predict how the maximum bending moment would change if the point load of $2\,400\ \text{N}$ were moved to the central section of the beam ($x = 3\ \text{m}$), justifying whether this new position is more or less unfavorable for the component.
A B P = 2400 N 2 m 4 m V ? ? M Mmax = ?
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Simply supported beam with a uniformly distributed load

Simply supported beam with a uniformly distributed load q, with its shear force diagram V (straight line passing through zero at the centre) and its bending moment diagram M (parabola with its maximum at the centre) q (N/m) A B L V +qL/2 −qL/2 V = 0 M Mmax = qL²/8 M = 0 M = 0
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A simply supported beam of $4\ \text{m}$ carries a uniformly distributed load of $500\ \text{N/m}$ along its entire length. a) Calculate the reactions. b) Calculate the…

A simply supported beam of $4\ \text{m}$ carries a uniformly distributed load of $500\ \text{N/m}$ along its entire length. a) Calculate the reactions. b) Calculate the maximum moment at the centre using $M_{max} = \frac{q \cdot L^2}{8}$. c) Draw the shape of the shear force diagram (an inclined line crossing zero) and the bending moment diagram (a parabola), marking their key values.

You can do it on paper and send a photo with your answer.

Solve
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A simply supported beam of $6\ \text{m}$ supports a uniform continuous load $q = 200\ \text{N/m}$ due to its self-weight. To serve an industrial facility, a stationary…

A simply supported beam of $6\ \text{m}$ supports a uniform continuous load $q = 200\ \text{N/m}$ due to its self-weight. To serve an industrial facility, a stationary machine of $1\,200\ \text{N}$ must be placed, and two alternatives are proposed: the central position A ($x = 3\ \text{m}$) or the lateral position B ($x = 1\ \text{m}$ from the left support).

  1. a) Applying the principle of superposition, obtain and quantitatively justify the value of the maximum bending moment for position A ($x = 3\ \text{m}$). Determine the percentage increase in the maximum bending moment in the span compared to the case where the beam only supported its self-weight.
  2. b) Compare the two alternatives with respect to the shear forces at the supports: predict and explain in which of the two situations the most heavily loaded support suffers more, and calculate the value of its reaction in alternative B ($x = 1\ \text{m}$).
  3. c) Justify under what mechanical conditions of the material and deformations it is physically valid to apply the principle of superposition in structural analysis, and state a specific example where this principle ceases to be applicable.
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The cantilever: maximum at the fixed support

Fixed cantilever beam with a vertical load at the free end and diagrams of constant shear force and linear bending moment, greatest at the fixed support P V constant M maximum at the fixed support
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Compare a simply supported beam with a cantilever beam from the perspective of their internal forces: in which one a positive moment appears at mid-span and in which one…

Compare a simply supported beam with a cantilever beam from the perspective of their internal forces: in which one a positive moment appears at mid-span and in which one a negative moment appears next to the support. Give a real example of each structural type (floor slab between joists, balcony) and indicate on which face the material works in tension in each case.

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A cantilever shelf of $1{,}5\ \text{m}$, fixed into the wall, carries $800\ \text{N}$ at its free end. a) Calculate the vertical reaction and the fixing moment at the…

A cantilever shelf of $1{,}5\ \text{m}$, fixed into the wall, carries $800\ \text{N}$ at its free end. a) Calculate the vertical reaction and the fixing moment at the wall. b) Draw the shear force diagram (constant) and the bending moment diagram (triangle with its maximum at the wall). c) Explain why the cantilever bracket would break at the fixed support rather than at the free end.

You can do it on paper and send a photo with your answer.

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A cantilever canopy of $2\ \text{m}$ supports a distributed load of $300\ \text{N/m}$ (self-weight plus snow). a) Replace the load with its resultant and calculate the…

A cantilever canopy of $2\ \text{m}$ supports a distributed load of $300\ \text{N/m}$ (self-weight plus snow). a) Replace the load with its resultant and calculate the moment at the fixed support. b) Check the result using the formula for maximum moment, $M_{max} = \frac{q \cdot L^2}{2}$. c) Draw the shape of the bending moment diagram (a parabola increasing towards the wall) and indicate its sign.

You can do it on paper and send a photo with your answer.

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The method of joints for trusses

Pin-jointed triangular truss with a vertical load at the top joint; support joint A, on the left, connects only two members with unknown forces and is therefore the starting point of the method of joints load joint A: 2 members unknowns: 2
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Describe what the method of joints for solving a truss consists of: what equilibrium equations are set up at each joint ($\sum F_x = 0$ and $\sum F_y = 0$), why it is…

Describe what the method of joints for solving a truss consists of: what equilibrium equations are set up at each joint ($\sum F_x = 0$ and $\sum F_y = 0$), why it is convenient to start with a joint that has only two unknown members, and how the calculation propagates to the rest of the structure.

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At a joint of a truss, a horizontal member and an inclined member sloping upwards to the left at $60^\circ$ to the horizontal meet, and a load of $1\,000\ \text{N}$…

At a joint of a truss, a horizontal member and an inclined member sloping upwards to the left at $60^\circ$ to the horizontal meet, and a load of $1\,000\ \text{N}$ hangs from it. a) Set up $\sum F_y = 0$ at the joint and calculate the force in the inclined member, explaining whether it is in tension. b) Set up $\sum F_x = 0$ and calculate the force in the horizontal member, explaining its sign (you may use $\sen 60^\circ = 0{,}866$ and $\cos 60^\circ = 0{,}5$).

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In the symmetrical roof truss with a span of $6\ \text{m}$ and a height of $2\ \text{m}$ shown in the figure, a load of $5\,000\ \text{N}$ is suspended from the bottom…

In the symmetrical roof truss with a span of $6\ \text{m}$ and a height of $2\ \text{m}$ shown in the figure, a load of $5\,000\ \text{N}$ is suspended from the bottom joint D:

  1. a) Set up the method of joints to calculate the axial forces in members AB, AD, and BD. Rigorously identify which are in tension and which are in compression.
  2. b) The diagonal compression member AB has a length of $3{,}61\ \text{m}$. Explain why, in the structural design of pin-jointed trusses, members under compression are much more critical and limiting than those under tension, naming the responsible physical phenomenon and justifying which geometric characteristics of their cross-section should be maximized to prevent it.
  3. c) Predict what would happen to the force distribution if the point load of $5\,000\ \text{N}$ were moved vertically from the bottom joint D to the top joint B, acting downward. Analytically prove what value the axial force in post BD would take and explain whether the forces in the perimeter members AB, BC, AD, and CD would increase, decrease, or remain identical.
A B C D ? ? ? P = 5000 N 3 m 3 m 2 m
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Summary

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