Analysis of Statically Determinate Structures

Learning Objectives

  • Define and calculate internal forces (normal, shear, and bending moment) within statically determinate beams and frames.
  • Apply standard sign conventions for drawing accurate shear and bending moment diagrams.
  • Utilize the differential relationships between load, shear, and moment to analyze structural behavior.
  • Determine the support reactions and internal forces for rigid frames using the equations of static equilibrium.
  • Evaluate the geometric stability and statical determinacy of structures, including those with internal hinges.
Methods for analyzing beams and frames under various loading conditions, focusing on internal forces and their distribution.

Beams

Beam

A structural member designed to support transverse loads. It transfers these loads to its supports through shear and bending moments. Common types include simply supported, cantilever, and continuous beams.

Internal Forces in Beams

Types of Internal Forces

When a beam is loaded, three internal forces develop at any cross-section to maintain equilibrium.

  • Normal Force (N): The axial force acting perpendicular to the cross-section.
  • Shear Force (V): The internal force acting parallel to the cross-section, tending to slide one part of the structure past the other.
  • Bending Moment (M): The internal moment acting about the neutral axis of the cross-section, tending to bend the structure.

Standard Sign Convention

Internal Forces Sign Convention

To ensure consistency when drawing shear and bending moment diagrams, a standard sign convention is universally adopted in structural engineering:

  • Axial Force (N): Positive when it creates tension (pulling away from the joint or section); negative when it creates compression.
  • Shear Force (V): Positive when the internal shear force on the left face of a cut section acts upwards, and the shear on the right face acts downwards. This tends to rotate the segment clockwise.
  • Bending Moment (M): Positive when it causes compression in the top fibers and tension in the bottom fibers of a beam (often remembered as causing the beam to "smile" or hold water). Negative moment causes tension in the top fibers.

Sign Convention Traps

A common mistake is forgetting that standard sign conventions for internal forces are different from the sign conventions used for global equilibrium (where usually right/up is positive). Always follow the standard internal force sign convention to avoid drawing flipped diagrams.

Principle of Superposition

Superposition

The Principle of Superposition states that for a linear-elastic structure, the total effect (e.g., deflection, shear force, bending moment) caused by multiple loads acting simultaneously is equal to the algebraic sum of the effects caused by each load acting individually.

  • Applicability: It is valid only when the structural material behaves linearly (Hooke's Law applies) and when the deformations are small enough that they do not significantly alter the geometry of the structure.
  • Utility: This principle greatly simplifies the analysis of complex loading scenarios by breaking them down into simpler, standard cases whose solutions are readily available in design tables.

Relationships Between Load, Shear, and Moment

Differential Relationship Between Load and Shear

The rate of change of shear force is equal to the negative of the distributed load intensity.

dVdx=โˆ’w(x)\frac{dV}{dx} = -w(x)

Variables

SymbolDescriptionUnit
VVShear forcekN
xxPosition along the beamm
w(x)w(x)Distributed load intensitykN/m

Differential Relationship Between Shear and Moment

The rate of change of bending moment is equal to the shear force.

dMdx=V(x)\frac{dM}{dx} = V(x)

Variables

SymbolDescriptionUnit
MMBending momentkNยทm
xxPosition along the beamm
V(x)V(x)Shear force at position xkN

Beam Analysis Procedure

  1. Reactions: Determine all external support reactions using the global equations of equilibrium (โˆ‘Fx=0\sum F_x = 0, โˆ‘Fy=0\sum F_y = 0, โˆ‘M=0\sum M = 0).
  2. Shear Diagram: Starting from the left end, plot the shear values. Concentrated forces cause sudden jumps in the shear diagram. Distributed loads cause a gradual change (slope equals the load intensity). The change in shear between two points equals the area under the load diagram.
  3. Moment Diagram: Starting from the left end, plot the moment values. Concentrated moments cause sudden jumps. The change in moment between two points equals the area under the shear diagram between those points. The maximum or minimum moment occurs where the shear is zero.

Frames

Frame

Structures composed of members connected at their ends by rigid joints. These rigid connections allow the transfer of bending moments, shear forces, and axial forces between members. Frames are common in building construction and industrial structures.

Analysis Approach for Frames

Frame Analysis Procedure

  1. Determine the support reactions by treating the entire frame as a single rigid body and applying the equations of equilibrium (โˆ‘Fx=0\sum F_x = 0, โˆ‘Fy=0\sum F_y = 0, โˆ‘M=0\sum M = 0).
  2. If the frame has internal hinges or pins, disassemble it at those points and apply equilibrium to the individual parts.
  3. Determine the internal forces (axial, shear, and moment) at specific locations or develop equations for them as functions of position along each member.
  4. Draw the axial force, shear force, and bending moment diagrams for each member.

Interactive Tool: Shear and Moment Simulation

Interactive Simulation

Visualize shear and moment diagrams for different beam loading configurations, such as point loads, uniform distributed loads, and varying distributed loads.

Beam Shear and Moment Simulator

Visualize internal forces for a simply supported beam (L=10textmL = 10 \\text{ m}).

Free Body Diagram

50 kN25.0 kN25.0 kN

Shear Diagram (V)Max: 25.0 kN

25.0-25

Bending Moment Diagram (M)Max: 125.0 kNยทm

125.0

Parameters

Position (x)5 m
Magnitude50 kN

Equations of Static Equilibrium

Global and Local Equilibrium

A statically determinate structure is one where all support reactions and internal forces can be found using only the equations of static equilibrium.

For a planar structure, the three equations of equilibrium are:

  • โˆ‘Fx=0\sum F_x = 0
  • โˆ‘Fy=0\sum F_y = 0
  • โˆ‘Mz=0\sum M_z = 0

If the number of unknown reactions equals 3 (in 2D), the structure is statically determinate externally. If unknown internal forces can be found without considering member deformations, it is statically determinate internally.

Determinacy and Stability Criteria

Geometric Instability

Even if a structure has enough reaction components, it can still be unstable if the reactions are improperly arranged. Examples of geometric instability include:

  • Parallel Reactions: All reaction lines of action are parallel (e.g., three roller supports). The structure cannot resist a lateral load.
  • Concurrent Reactions: All reaction lines of action intersect at a single point. The structure will rotate about that point under a general load.

Internal Hinge Condition Equations

Internal hinges (or pins) provide additional equations of equilibrium because they cannot transmit bending moments. For a structure with nn internal hinges, nn additional equations of the form โˆ‘Mhinge=0\sum M_{\text{hinge}} = 0 can be written by considering the free-body diagram of the structure on either side of the hinge. This allows for the analysis of otherwise seemingly indeterminate structures (e.g., a three-hinged arch or a Gerber beam).

Key Takeaways
  • Beams support transverse loads through internal shear and bending moments.
  • Internal forces in beams include axial force, shear force, and bending moment.
  • The relationships between load, shear, and moment are given by dV/dx=โˆ’w(x)dV/dx = -w(x) and dM/dx=V(x)dM/dx = V(x).
  • The change in shear equals the area under the load diagram, and the change in moment equals the area under the shear diagram.
  • Frames consist of rigidly connected members that transmit axial, shear, and moment forces.
  • The analysis of frames involves finding support reactions, internal forces at joints, and plotting diagrams for each member.
  • Statically determinate structures can be completely analyzed using only equations of equilibrium.
  • Stability requires adequate support conditions to prevent rigid-body motion.
  • Internal hinges provide essential condition equations (โˆ‘M=0\sum M = 0) that allow statical determinacy for specific structural forms.