Structural Mechanics
2-Force Members
A 2-force member is a structural component subjected to only two forces at its endpoints. These forces must be equal in magnitude, opposite in direction, and collinear in order to maintain equilibrium. A 2-force member can only experience tension (pulling apart) or compression (pushing together) along its length. No bending, shear, or torsional forces act on it.
In equilibrium, the forces must act along the axis of the member. If the forces attempt to act in any other direction or introduce moments, the member will not remain stable. Therefore, 2-force members are limited to pure tension or compression to maintain equilibrium.
Consider a rod in 2D, subjected to two forces \mathbf F_1 and \mathbf F_2, applied at its endpoints. The rod is a 2-force member, and for it to remain in equilibrium, these forces must be aligned with the rod’s axis. Let’s show that the tangential (perpendicular to the rod’s axis) component of the forces must be zero.
- Length of the rod: L
- Forces: \mathbf F_1 at one end, \mathbf F_2 at the other
- Assume the rod is aligned along the x-axis, it is always possible to rotate the axis to arrive in this configuration.
The forces can be broken into components:
\begin{aligned} \mathbf F_1 & = (F_{1x}, F_{1y}) \\ \mathbf F_2 & = (F_{2x}, F_{2y}) \end{aligned}
For the rod to be in equilibrium, both the sum of forces and the sum of moments about any point must be zero.
For the equilibrium of force:
\begin{aligned} & \sum \mathbf F_x = F_{1x} + F_{2x} = 0 \\ & \sum \mathbf F_y = F_{1y} + F_{2y} = 0 \end{aligned}
For the equilibrium of moments should be true about any point, for example the left endpoint of the rod (where \mathbf F_1 acts):
M = L \cdot F_{2y}
For rotational equilibrium:
M = 0 \quad \Rightarrow \quad L \cdot F_{2y} = 0
Since L \neq 0, we must have:
F_{2y} = 0
Then applying the equilibrium of forces along y:
F_{1y} = 0
Both F_{1y} and F_{2y}, the tangential forces perpendicular to the rod, must be zero. Therefore, the forces must act purely along the rod’s axis (either in compression or tension), and the tangential force is zero for equilibrium in a 2-force member.
The shape is not a factor to be considered as far there are only 2 forces, they act on the line of action connecting the points where these forces are applied.
Multi-force members
A multi-force member is a structural component subjected to more than two forces at different points along its length. Unlike 2-force members, where the forces must act along the member’s axis in tension or compression, multi-force members experience a combination of forces, which may include tension, compression, bending, shear, and torsion. These forces are typically non-collinear, and their equilibrium is determined by the sum of forces and moments being zero. Multi-force members are common in complex structures like beams and frames, where forces are applied at multiple locations, resulting in internal stress distributions.
In multi-force systems, the number of unknowns typically exceeds the number of available equilibrium equations (three in 2D: two for forces and one for moments). To solve for the unknowns, we apply the principle of subdivision into parts:
If a body is in equilibrium, each part of the body, when considered separately, must also be in equilibrium.
By analyzing smaller sections, internal forces and reactions can be exposed, leading to additional equations
Frames are rigid structures composed of multiple interconnected members, designed primarily to support loads and remain stationary. They are often used in buildings, bridges, and towers to provide stability and strength. In a frame, the forces applied to the structure are transferred through its members, which may experience a combination of tension, compression, bending, and shear forces. Each member in the frame can be a multi-force member, subjected to forces at various points. A common example of a frame is the structural skeleton of a building, where the columns and beams support the weight of the structure and any external loads.
Machines, on the other hand, are mechanical systems that include moving parts and are designed to transmit motion and forces to perform work. Machines often consist of multiple interconnected members, such as links, gears, and levers, that move relative to each other. These members experience various forces, including tension, compression, bending, and torsion, depending on the motion and external loads applied. A typical example of a machine is a crane, which uses levers and pulleys to lift and move heavy loads. Other examples include engines, robotic arms, and mechanical linkages used in manufacturing. Unlike frames, machines are characterized by their ability to perform tasks through motion..
Trusses
A truss is a structural framework composed of straight slender members connected at their ends, typically in triangular units, designed to support loads. The connections, or joints, are usually assumed to be pin joints, which allows the members to carry only axial forces (tension or compression) and not bending moments, therefore by construction all members in a truss are 2-force members.
Trusses offer several advantages, such as providing high strength while allowing for long spans with minimal material weight. They efficiently distribute loads, making them ideal for large structures like bridges and roofs. The use of triangular configurations enhances their stability, and because they use less material compared to solid beams, they are cost-effective and reduce the overall weight of the structure without compromising strength.
The assumptions I will make are that the loading is applied exclusively at the joints, the pins are frictionless, and the weight of the members is negligible compared to the applied forces. These simplifications allow for a more straightforward analysis, where the truss members are subjected purely to axial forces, ensuring that they experience either tension or compression without any bending moments or shear forces.
The approach I will take is to cut through the member, not through the joint, in order to isolate the portion of the truss that interests me. I will always assume the forces in the members are in tension, and if the result is negative, it indicates compression. External reactions will only be solved if they fall within the portion of the truss isolated by the cut.
The method of joints is a technique used to solve for the forces in the members of a truss. The key idea is to isolate each joint, treating it as a point where all the forces meet. Since the forces all act along the lines of the members and intersect at the joint, we can ignore moments (they are zero) because the forces do not create any rotational effect. For each joint, we solve for two unknown forces by applying the conditions of static equilibrium: the sum of the forces in both the horizontal and vertical directions must be zero. This method is particularly useful for trusses with two unknown forces at each joint, allowing for systematic solving of the truss one joint at a time.
Another method that can be used is the method of sections. Unlike the method of joints, where forces are concurrent at a single point, in the method of sections we isolate a section of the truss that includes more than one joint. Since the forces are no longer concurrent, this method requires considering not just the forces, but also the moments acting on the isolated section. By applying the conditions of static equilibrium—sum of forces in both horizontal and vertical directions equal to zero, and the sum of moments equal to zero—we can solve for the forces in several members at once. This approach is especially useful when targeting specific members for analysis without needing to solve the entire truss step-by-step.
Zero force members
A zero-force member in a truss is a structural component that, under certain loading conditions, carries no load. These members typically occur in situations where joints are unconnected to external forces or symmetric configurations lead to no force being transmitted through specific members.
Zero-force members are useful because they provide stability during construction, help maintain the overall geometry of the truss, and can be important under different loading scenarios, even though they may not carry any load in the current configuration. Their presence ensures that the structure can respond effectively to unexpected changes in load conditions.
Finally, in some cases are inserted to give a better aesthetic look at the overall structure.
In addition to perform a full analysis, there is a visual way to identify zero-force members in a truss based on the arrangement of joints and members:
- if a joint has only two non-collinear members and no external load or support reaction is applied at that joint, both of those members are zero-force members.
- if a joint has three members, two of which are collinear, and there is no external load or reaction at that joint, the non-collinear member is a zero-force member. These visual rules can be applied to quickly identify zero-force members during analysis and greatly simplify the computation.
Cable systems
Cable structures are systems where cables, acting as tension members, are the primary load-bearing elements. These structures are highly efficient in resisting tensile forces but cannot support compression or bending. The flexibility of cables allows them to adapt to various loading conditions. A key characteristic of cable structures is that the shape of the cable changes with the type and magnitude of the applied load.
Examples of cable structures include different categories based on how they handle loads:
-
For self-weight, an example would be power transmission lines. These cables carry only their own weight, and the shape they form under this condition is typically a catenary.
-
In the case of concentrated loads, an example is a cable carrying a set of discrete loads, such as an elevator cable where the weight is concentrated at specific points along the cable’s length.
-
Suspension bridges are an example of cable structures that support distributed loads. Here, the cables carry the weight of the bridge deck and any additional traffic loads. The shape of the cables in a suspension bridge under a uniform load is a parabola.
Cable structures offer several advantages. They are highly efficient in supporting tensile forces, allowing for the construction of large-span structures with minimal material use. This leads to lighter structures and reduced construction costs.
Their flexibility allows them to adapt to different loading conditions without significant deformation, making them suitable for dynamic loads like wind and earthquakes. Cable structures also enable elegant and aesthetically pleasing designs, as seen in many iconic bridges and roofs.
Additionally, the use of cables provides structural redundancy, meaning the failure of one cable does not necessarily lead to the collapse of the entire system.
Concentrated load
In cable systems subjected to concentrated loads, the cables experience tension forces that adjust based on the locations of the loads. These systems, assuming pin connections, allow only tensile forces without any moments, both external and internal, making the cables perfectly flexible and incapable of resisting bending or compression. Additionally, the assumption of inextensibility means the length of the cable does not change under loading.
Examples of such systems include:
For a traffic light hung by cables, the concentrated load comes from the weight of the traffic light itself. The cables simply adjust their tension to maintain equilibrium.
In an aerial tram or ski lift, the cable carries the weight of the cabins or gondolas at discrete points. These loads are applied at specific intervals along the cable, and the tension in the cable adjusts to support both the weight of the vehicles and any dynamic forces from movement.
These systems rely entirely on the tensile capacity of the cables to support the loads without any bending resistance, which is why they can span significant distances with minimal material usage, while maintaining flexibility.
Uniform load
In a cable system, the curve formed by a cable under its own weight is a catenary, while the curve formed by a cable under a uniform distributed load is a parabola. The difference between these curves lies in the nature of the forces acting on the cable and how they influence its shape.
The catenary equation for a cable subjected to its own weight is given by:
y(x) = a \cosh{\left(\frac{x}{a}\right)} - a
where a is a constant related to the tension in the cable, x is the horizontal distance along the cable, and y(x) is the vertical displacement at any point. The shape of this curve is seen in structures like suspension bridges and power lines, where the self-weight of the cable dominates. For example, long spans of high-voltage power lines exhibit a near-perfect catenary because their primary load is their own weight.
On the other hand, when a cable is subjected to a uniform distributed load, such as a roadbed in the case of suspension bridges, the cable takes on a parabolic shape. The equation for the parabolic curve under uniform loading is:
y(x) = \frac{q x^2}{2T_0}
where q is the uniform load per unit length, T_0 is the horizontal tension at the lowest point of the cable, and x and y(x) represent the horizontal and vertical coordinates, respectively. In the case of suspension bridges, such as the Golden Gate Bridge or Brooklyn Bridge, the load from the roadway is evenly distributed, leading to a parabolic shape in the main cables, even though the cables themselves may approximate a catenary when considering self-weight alone.
Internal force in beams
When A beam is loaded, there are force acting internally on the material.
Shear force at a section in a beam is the internal force acting perpendicular to the cross-section, resulting from the applied external forces, which causes one part of the beam to slide relative to the other.
Bending moment at a section in a beam is the internal moment that causes the beam to bend, resulting from external forces or loads applied to the beam. It is the rotational force acting about the section, causing a curvature.
Let’s consider a cantilever beam which has a force F applied on one edge; the reaction at the wall will create some force internally on the beam itself, and lets assume to cut it through a section.
Considering first the right part, this section still need to be in equilibrium; applying the equations of equilibrium, in the vertical direction a shear force V must be applied, and also an additional moment M is present to balance a rotation that would be induced by the force F.
Considering the other section of the beam, there will be an equal and opposite force V and momentum M that are applied.
The sign convention for shear force and bending moments in beams is as follows. For shear force, if the force causes a clockwise rotation on the material of the beam, it is considered positive. If it causes a counterclockwise rotation, the force is negative.
For bending moments, a moment is considered positive if it results in a concave-up curvature of the beam, often described as a “smiley face,” and negative if it causes a concave-down curvature, resembling a “frown.” This convention helps in consistently determining the internal forces and moments throughout the beam’s length.
For a beam subjected to a distributed load q(x), the differential relationships between the load, shear force V(x), and bending moment M(x) are as follows.
The differential equation relating the distributed load and the shear force is:
\frac{\mathrm dV(x)}{\mathrm dx} = -q(x)
This indicates that the rate of change of the shear force along the beam is equal to the negative of the distributed load at that point.
The differential equation relating the shear force and the bending moment is:
\frac{\mathrm dM(x)}{\mathrm dx} = V(x)
This shows that the rate of change of the bending moment is equal to the shear force.
For the integral form, integrating the differential relationships provides:
V(x) = V(x_0) - \int_{x_0}^{x} q(\xi) \, \mathrm d\xi
This gives the shear force at any point x as a function of the initial shear force V(x_0) and the integral of the distributed load over the distance from x_0 to x.
Similarly, the bending moment is given by:
M(x) = M(x_0) + \int_{x_0}^{x} V(\xi) \, \mathrm d\xi
This expresses the bending moment at any point x in terms of the initial moment M(x_0) and the integral of the shear force over the distance from x_0 to x.