CELE Engineering Mechanics — Frames, Machines and CablesSummary

A frame is a stationary structure with at least one multi-force member (a member loaded at more than two points or carrying an applied moment), producing shear and bending in addition to axial force. A truss contains only two-force members and is idealized to carry only axial force. A machine is similar to a frame but designed to transmit or modify forces (e.g., pliers, jacks, linkages) and may contain moving or sliding elements. All are analyzed by dismembering into free-body diagrams and applying equilibrium; the key difference is the type of forces each member develops and the purpose of the structure.

Step 1: Find external reactions from the FBD of the entire structure (sum of forces and moments). Step 2: Draw a separate FBD for each member or group of members, showing internal pin forces at joints as action-reaction pairs (Newton's third law — equal in magnitude, opposite in direction on the two adjoining members). Step 3: Apply equilibrium equations (ΣFₓ = 0, ΣFᵧ = 0, ΣM = 0) to each FBD to solve for unknown forces. Two-force members should be identified first, since their internal force acts along the member axis, immediately giving direction and reducing unknowns.

A two-force member is loaded only at its two ends (typically at pins). For equilibrium, the internal force must act along the line joining the two pins. This member acts as a strut (compression) or tie (tension). Identifying two-force members instantly provides direction of the internal force, eliminating one unknown (direction) and simplifying the dismembering process. For example, a diagonal strut in a frame, if pinned only at its ends with no transverse load, carries force along its axis.

A perfectly flexible cable carries only tension, directed tangent to the cable at any point. For a cable under load, the horizontal component of tension H is constant throughout the length (found from horizontal equilibrium). The vertical component varies with load. At any point, the total tension T is found from T = H / cos(θ), where θ is the angle of the cable with the horizontal. Tension is maximum at the steepest point (typically at the supports where the cable rises most steeply), not at midspan.

When a cable supports a uniform load distributed over the horizontal span (e.g., the deck of a suspension bridge), the cable forms a parabola. For a span L, uniform load w (per unit horizontal distance), and sag d at midspan: Horizontal component H = wL²/(8d); Maximum tension at supports T_max = √[H² + (wL/2)²]. The parabolic shape is also a good approximation for catenary cables with shallow sag (d/L < 0.1) where self-weight dominates the distributed load.

When a cable is loaded primarily by its own weight (self-weight dominates over applied loads), the curve is a catenary, following y = (H/w)[cosh(wx/H) - 1], where H is the horizontal component and w is the weight per unit length of cable. The catenary is more deeply curved than a parabola for the same span and sag. For shallow sags (d/L < 0.1), the parabolic approximation is acceptable and simpler to use; for deep sags, the catenary formula must be employed.

A cable carrying concentrated loads at discrete points forms a funicular polygon (straight-line segments between loads). The horizontal component H remains constant throughout. At each load point, the cable changes slope. The tension in each segment is T = H / cos(θ), where θ is the segment's angle. To find H, apply vertical equilibrium at the load point: ΣFᵧ = 0 relates the vertical components of tension in adjacent segments to the applied load.

Mechanical advantage (M.A.) is the ratio of the output load to the input effort, or equivalently, the ratio of the effort arm to the load arm (for a lever): M.A. = Load / Effort = Effort Arm / Load Arm. An M.A. > 1 means the machine amplifies force (but reduces displacement). An M.A. < 1 means the machine trades force for distance. Ideal M.A. (ignoring friction) is found by geometry; actual M.A. is reduced by friction and is always less than ideal.

At a pin joint connecting multiple members, all forces (from the members and any external load at the pin) must sum to zero (ΣFₓ = 0, ΣFᵧ = 0). When dismembering, the pin force on one member is drawn equal and opposite to the pin force on the other member (Newton's third law). For multi-force members, internal shear and bending develop; the pin force is one of many forces, and moment equilibrium (ΣM = 0) is required to solve for the pin force fully.

A multi-force member carries transverse loads, applied moments, or forces at more than two points. These loads produce both axial and transverse forces (shear) and bending moments within the member. Dismembering and equilibrium allow us to find the internal shear and moment as functions of position along the member. Understanding shear and bending is foundational to stress analysis, deflection, and design per NSCP 2015 and AISC 360.

The formula H = wL²/(8d) shows that for fixed span L and load w, increasing sag d decreases the required horizontal component H (and thus the support reactions and material stress). Conversely, for fixed load and span, reducing sag requires stronger cables. This trade-off is central to suspension bridge design.