Statically Determinate Beams and Frames

Introduction

Beams and frames are fundamental structural systems in buildings, bridges, temporary works, and construction equipment. A structural analysis determines how externally applied loads are transferred through members to supports and foundations. For a statically determinate structure, all support reactions and internal actions can be obtained using the equations of static equilibrium alone.

The principal internal actions are axial force, shear force, and bending moment. Their variation along a member is represented by axial-force, shear-force, and bending-moment diagrams. These diagrams allow engineers to identify critical sections, select member sizes, check connections, and understand how a structure carries load.

This chapter considers idealized planar beams, compound beams, and rigid frames subjected primarily to concentrated forces, distributed loads, and applied couples.

Learning Objectives

By the end of this chapter, a student should be able to:

1. Identify common supports and determine their reaction components.
2. Assess whether a planar structure is statically determinate.
3. Calculate support reactions using equilibrium equations.
4. Determine internal axial force, shear force, and bending moment.
5. Construct and interpret force diagrams.
6. Analyze compound beams and rigid frames.
7. Apply equilibrium principles to practical construction situations.
8. Recognize assumptions, sign conventions, and limitations of idealized analysis.

1. Structural Idealization and Determinacy

A real structure contains three-dimensional members, flexible connections, material imperfections, and foundation movements. In elementary structural analysis, it is represented by an idealized model consisting of loads, members, joints, and supports. The model must preserve the important load-transfer behavior while simplifying details that do not significantly affect the required calculation.

A planar structure is in static equilibrium when the resultant force and resultant moment are zero. For a body acted upon in a plane, the independent equilibrium equations are:

Here, F_x and F_y are force components in the horizontal and vertical directions, and M_O is the algebraic moment about any point O. Moments are commonly measured in kN·m, while forces are measured in kN.

For a force F acting with perpendicular distance d from a point, the moment magnitude is:

The equation applies when d is the perpendicular distance from the moment center to the force line of action. Counterclockwise moments are often taken as positive and clockwise moments as negative, although any consistent convention is acceptable.

Common supports

A structure is statically determinate when all unknown reactions and internal forces can be found from independent equilibrium equations. For a single planar free body, three independent equilibrium equations are available. If the number of independent unknown reactions exceeds the available equations, the structure is statically indeterminate. If fewer reactions exist, the structure may be unstable.

Determinacy must be assessed using structural connectivity as well as reaction counting. For example, an internal hinge supplies an additional condition: the bending moment at that hinge is zero. It also permits separate free-body diagrams for the portions connected by the hinge.

2. Loads and Resultants

Structural loads may be concentrated, distributed, or applied as couples.

A point load acts at a specified location. A uniformly distributed load of intensity w kN/m acting over length L has resultant:

The resultant acts through the centroid of the loaded length. For a uniform load, this is at its midpoint. Thus, replacing a uniformly distributed load by a single force is valid for calculating overall reactions, provided its magnitude and line of action are preserved.

For a linearly varying load forming a triangular distribution, the resultant is:

It acts one-third of the base length from the larger-intensity end, or two-thirds from the zero-intensity end. These resultant-load rules apply to reaction calculations, but the original load distribution must be restored when constructing local shear and moment diagrams if the exact variation is required.

An applied couple has a specified moment independent of the point about which it is evaluated. It produces no net force, but it changes the moment equilibrium and causes a jump in the bending-moment diagram.

3. Support Reactions: General Procedure

A reliable reaction calculation follows these steps:

1. Draw the complete structural free-body diagram.
2. Replace each support by its reaction components.
3. Replace distributed loads by equivalent resultants when calculating overall reactions.
4. Select a sign convention.
5. Take moments about a point that eliminates as many unknowns as possible.
6. Use force equilibrium to determine the remaining reactions.
7. Check the result using an independent equilibrium equation.

A negative calculated reaction does not automatically indicate an error. It means the actual direction is opposite to the assumed direction. However, a support such as a roller that can only provide compression may lift off if the calculated reaction is physically incompatible with the support model.

Worked Example 1: Simply supported beam

A 6 m beam is supported by a pin at A and a roller at B. A 20 kN point load acts 2 m from A, and a uniformly distributed load of 4 kN/m acts over the full span. Determine the support reactions.

The distributed-load resultant is:

It acts 3 m from A. Let the pin reactions be A_x and A_y, and the roller reaction be B_y. There are no horizontal applied loads, so:

Taking moments about A:

Therefore:

Vertical equilibrium gives:

The reactions are A_x = 0, A_y = 25.33 kN upward, and B_y = 18.67 kN upward. Their sum, 44 kN, equals the total applied vertical load.

4. Internal Actions and Sign Conventions

To determine internal actions at a section, imagine cutting the member at that section and draw the free-body diagram of one side. The exposed internal actions are:

• Axial force N: force along the member axis.
• Shear force V: force perpendicular to the member axis.
• Bending moment M: internal couple resisting bending.

A common convention defines positive axial force as tension, positive shear according to the selected face convention, and positive bending moment as sagging. Other conventions are acceptable if used consistently. The essential requirement is that the signs in equations and diagrams agree with the stated convention.

For a beam with distributed load intensity w(x), shear V(x) and bending moment M(x) are related by:

These relationships assume a consistent coordinate direction and sign convention. They show that load controls the slope of the shear diagram, while shear controls the slope of the moment diagram. Consequently:

• A concentrated force causes a jump in the shear diagram.
• A distributed load produces a sloping shear diagram.
• A concentrated couple causes a jump in the moment diagram.
• Where shear is zero, bending moment has a stationary value, often a maximum or minimum.
• Where no distributed load acts, shear is constant and moment varies linearly.

At a free end with no applied end force, shear is zero. At a free end with no applied couple, moment is zero. At a simple support, the bending moment is normally zero unless an external couple is applied at that location.

5. Shear-Force and Bending-Moment Diagrams

A systematic diagram procedure is as follows:

1. Calculate reactions.
2. Mark all load discontinuities, supports, hinges, and member ends along the beam axis.
3. Determine shear values immediately to the left and right of each concentrated force.
4. Use the load intensity to establish the shape of the shear diagram.
5. Determine moment values at supports and key sections.
6. Use the area under the shear diagram to obtain changes in moment.
7. Check that the final boundary conditions are satisfied.

The change in moment between two sections is equal to the signed area under the shear diagram:

The units are consistent because shear has units of force and distance has units of length, giving force × length.

Worked Example 2: Point-loaded simply supported beam

A simply supported beam AB has span 8 m. A 30 kN point load acts at midspan. Determine the reactions and key values for the shear and bending-moment diagrams.

By symmetry:

The shear is +15 kN from A to the load. At the load it drops by 30 kN, becoming −15 kN until B, where the reaction increases it to zero.

For the left half, at a distance x from A:

At midspan:

For the right half, using distance x from A:

At B, M(8) = 0, as required for a simple support. The maximum moment occurs where shear changes sign, at the point load, and equals 60 kN·m.

For the beam in Example 1, the bending moment at the point load is also useful. On the left segment, immediately under the 20 kN load at x = 2 m:

The exact maximum moment occurs where shear is zero, not necessarily directly under the point load. For the left region, shear is:

Setting V = 0 gives x = 6.33 m, which lies to the right of the point load, so the maximum is located there. Substitution gives approximately:

This illustrates why locating a maximum moment from load position alone can be incorrect when several loads act together.

6. Compound Beams and Internal Hinges

A compound beam consists of two or more beam segments connected so that the whole system acts as one structure. If the connection is an internal hinge, it transmits axial and shear forces but no bending moment. Therefore:

The internal hinge provides a valuable additional condition. The usual method is:

1. Draw the complete free body and identify external reactions.
2. Isolate one segment at the internal hinge.
3. Use moment equilibrium about the hinge or another convenient point.
4. Determine hinge forces.
5. Analyze each segment for internal actions.

Illustrative compound-beam application

Consider a beam with a pin at A, a roller at B, and an internal hinge H between them. A load acts on the left segment and another load acts on the right segment. The hinge divides the structure into two free bodies. Since the moment at H is zero, the left segment can often be solved by taking moments about A and H, while the right segment is solved using the equal-and-opposite hinge forces.

The hinge forces must satisfy action–reaction: if the left segment experiences a horizontal force H_x to the right and vertical force H_y upward, the right segment experiences H_x to the left and H_y downward. Failure to reverse both components is a common source of error.

7. Rigid Frames

A rigid frame consists of members connected so that joints transfer forces and moments. Unlike a pin-jointed truss, a rigid frame may carry significant bending moments in its columns and beams. External reactions are first found from the free body of the entire frame. Internal actions are then calculated by cutting individual members or by isolating portions of the frame.

For a frame joint without an internal hinge, the connected members share force and moment interaction. At a joint, equilibrium requires:

where J is the joint center. A rigid connection can transmit a moment; a pin connection cannot.

Frame analysis procedure

1. Replace supports with appropriate reactions.
2. Treat the complete frame as one free body to calculate external reactions.
3. Select a section through the member of interest.
4. Draw the free body of one side of the cut.
5. Resolve forces into local axial and transverse directions.
6. Apply force and moment equilibrium.
7. Repeat at critical sections and construct diagrams member by member.

For an inclined member, global force components must be transformed into member axes. If a global force has components F_x and F_y, and the member axis makes angle θ with the horizontal, the axial component is:

The transverse component, using one common orientation, is:

The signs depend on the selected local axes. These equations are simply a coordinate transformation and require consistent angle and sign definitions.

Frame example: qualitative interpretation

A portal frame with fixed or pinned bases and a horizontal roof beam illustrates the different roles of axial force, shear, and moment. A vertical gravity load generally produces compression in columns and bending in the beam. A lateral wind load produces horizontal reactions, column shear, and significant bending moments at rigid joints. The largest moment may occur at a column base or beam-column connection rather than at midspan.

A frame diagram should therefore be interpreted member by member. A zero moment at a pin support does not mean that the adjacent rigid joint also has zero moment. Similarly, an internal force diagram can change abruptly at a concentrated joint force or applied couple.

8. Detailed Applied Case: Temporary Construction Platform

Background

Consider an illustrative, non-project-specific case involving a temporary steel platform supporting stored masonry units. The platform is idealized as a simply supported beam spanning 5 m. A 12 kN concentrated load from a pallet acts 1.5 m from the left support. A uniformly distributed self-weight and deck load of 2 kN/m acts over the entire span.

Problem

Determine the support reactions and identify the critical bending moment location for preliminary member assessment.

Analysis

The distributed-load resultant is:

It acts 2.5 m from the left support. Let reactions be A_y and B_y. Taking moments about A:

Vertical equilibrium gives:

For a section to the left of the pallet load, the shear is:

Setting V = 0 gives x = 6.7 m, outside this region. To the right of the pallet load:

Setting this equal to zero gives:

This is not in the region to the right of the pallet, which begins at 1.5 m. Thus the shear changes from positive to negative at the 12 kN load itself, so the maximum bending moment occurs at x = 1.5 m.

The moment there is:

Outcome and professional interpretation

The left support carries 13.4 kN and the right support 8.6 kN. The critical positive moment is approximately 17.85 kN·m at the pallet location. A design engineer would not stop at this calculation: member resistance, lateral stability, bearing, connection capacity, vibration, load placement, construction-stage instability, and applicable temporary-works regulations must also be checked.

Learning points

• Load position strongly influences reactions and moment.
• A maximum moment occurs where shear changes sign or at a discontinuity where the sign changes across a point load.
• An analysis model is not a substitute for checking actual load paths and site conditions.
• Temporary structures require particular attention to accidental loading, incomplete bracing, and erection sequence.

9. Quality Checks, Limitations, and Professional Practice

Useful checks include:

• The sum of vertical reactions equals the total applied vertical load.
• The sum of horizontal reactions equals the total applied horizontal load.
• The sum of moments about a second point is zero.
• A simple support has the expected moment condition.
• The shear diagram returns to zero after all forces are included.
• The moment diagram satisfies end conditions.
• Units remain consistent throughout.

The methods in this chapter assume small deformations, idealized supports, known geometry, static loading, and planar behavior. They do not by themselves address buckling, dynamic amplification, material nonlinearity, three-dimensional torsion, soil–structure interaction, fatigue, or plastic collapse. Real design also requires load combinations, safety factors, material standards, connection design, fire considerations, and relevant building codes.

Computer structural-analysis software can efficiently produce force diagrams for complex models, but software output must be verified. A professional should inspect the model, support conditions, load directions, releases, units, mesh or element idealization, and equilibrium residuals. Artificial intelligence may assist with model documentation or preliminary interpretation, but it must not replace engineering judgment, independent checking, or the responsibility of a qualified engineer.

Best-Practice Workflow

1. Sketch the real load path before writing equations.
2. Clearly label dimensions, loads, supports, and axes.
3. Use a complete free-body diagram before isolating parts.
4. Replace distributed loads by resultants only when their line of action is preserved.
5. State the sign convention beside every diagram.
6. Use equilibrium at more than one stage as an error check.
7. Locate extrema by examining shear, not by assuming they occur at midspan.
8. Distinguish external reactions from internal actions.
9. Record units with numerical results.
10. Compare analytical results with physical intuition and, where necessary, an independent computational model.

Chapter Summary

Statically determinate beams and frames can be analyzed using the three planar equilibrium equations: horizontal force equilibrium, vertical force equilibrium, and moment equilibrium. Support idealization determines the unknown reaction components and whether a structure is stable and determinate.

Distributed loads are replaced by equivalent resultants for overall equilibrium, with the resultant acting through the load distribution’s centroid. After reactions are determined, internal axial force, shear, and bending moment are found by cutting the structure and applying equilibrium to a free body.

Shear and moment diagrams provide a visual description of internal actions. Concentrated forces cause jumps in shear, distributed loads create shear slopes, concentrated couples cause moment jumps, and the area under the shear diagram equals the change in moment. Maximum or minimum bending moments generally occur where shear is zero or changes sign.

Compound beams require careful treatment of internal hinges, which transmit forces but no bending moment. Rigid frames require member-by-member analysis because rigid joints transmit moments as well as forces. All results must be checked against equilibrium, boundary conditions, units, physical behavior, and applicable codes. Analytical methods remain essential even when computer software is used.

Key Terms

Static equilibrium: Condition in which the resultant force and resultant moment are zero.

Statically determinate: A structure whose unknown reactions and internal actions can be found from independent equilibrium equations.

Free-body diagram: Isolated representation of a body showing all external forces and moments acting on it.

Support reaction: Force or moment supplied by a support to maintain equilibrium.

Pin support: Support providing two force reaction components in a plane.

Roller support: Support providing one reaction normal to its rolling surface.

Fixed support: Support resisting two translations and rotation in a plane.

Internal hinge: Connection that transmits force but releases bending moment.

Axial force: Internal force acting along a member’s longitudinal axis.

Shear force: Internal force acting transverse to a member’s longitudinal axis.

Bending moment: Internal couple associated with flexural action.

Point load: Concentrated force applied at a specified location.

Distributed load: Load spread continuously over a length or area.

Resultant load: Single force equivalent to a distributed loading for equilibrium purposes.

Shear-force diagram: Graph showing shear-force variation along a member.

Bending-moment diagram: Graph showing bending-moment variation along a member.

Rigid frame: Structural assembly whose joints transmit forces and bending moments.

Sagging moment: Common positive bending condition producing concave-up curvature in a beam diagram convention.

Action–reaction: Equal and opposite interaction forces between connected bodies.

Load path: Physical route through which applied loads are transferred to foundations or supports.

Review Questions

Short Answer Questions

1. What three independent equilibrium equations apply to a planar rigid body?
2. Why is the resultant of a uniform distributed load placed at the midpoint of its loaded length?
3. What internal actions are exposed when a beam is cut at a section?
4. What is the bending moment at an ideal internal hinge?
5. What does a jump in a shear-force diagram represent?

Long Answer Questions

1. Explain the procedure for calculating reactions of a simply supported beam carrying point and distributed loads.
2. Derive the relationships between distributed load, shear force, and bending moment, and explain their diagrammatic meaning.
3. Describe how a compound beam with an internal hinge is analyzed using separate free-body diagrams.
4. Compare the behavior of a pin-jointed truss and a rigid frame under vertical and horizontal loading.
5. Explain the assumptions and limitations of statically determinate beam analysis in construction practice.

Analytical / Application Questions

1. A 7 m simply supported beam carries a 16 kN point load 2 m from the left support and a 3 kN/m uniform load over the full span. Calculate both reactions and identify the section of maximum bending moment.
2. A compound beam contains an internal hinge. Explain how the hinge condition can be used to determine unknown reactions and why hinge forces on the two isolated segments must be equal and opposite.
3. A temporary frame is analyzed using computer software, but the horizontal reaction is unexpectedly zero under wind loading. Identify at least four modeling or loading checks that should be performed before accepting the result.

Multiple Choice Questions with Answers

Question 1

Which support generally provides two force reactions in a planar structural model?

A. Roller support

B. Pin support

C. Free end

D. Internal hinge

Correct Answer: B. Pin support

Question 2

A uniformly distributed load of 5 kN/m acts over 4 m. What is its equivalent resultant?

A. 1.25 kN

B. 9 kN

C. 20 kN

D. 25 kN

Correct Answer: C. 20 kN

Question 3

For a planar rigid body in equilibrium, which condition must hold?

A. Only the vertical forces must sum to zero

B. Only the moments must sum to zero

C. The force resultant and moment resultant must both be zero

D. The horizontal reaction must be zero

Correct Answer: C. The force resultant and moment resultant must both be zero

Question 4

What does a concentrated transverse force normally cause in a shear-force diagram?

A. A vertical jump

B. A parabolic curve

C. A horizontal-axis intercept only

D. No change

Correct Answer: A. A vertical jump

Question 5

What is the bending moment at an ideal internal hinge?

A. Maximum positive moment

B. Equal to the shear force

C. Zero

D. Equal to the axial force

Correct Answer: C. Zero

Question 6

For a simply supported 8 m beam carrying a 30 kN point load at midspan, what is the maximum bending moment?

A. 30 kN·m

B. 60 kN·m

C. 120 kN·m

D. 240 kN·m

Correct Answer: B. 60 kN·m

Question 7

The slope of the bending-moment diagram at a section is equal to the:

A. Axial force

B. Distributed load

C. Shear force

D. Support reaction only

Correct Answer: C. Shear force

Question 8

A 10 kN point load acts 2 m from a moment center. Its moment magnitude is:

A. 5 kN·m

B. 12 kN·m

C. 20 kN·m

D. 40 kN·m

Correct Answer: C. 20 kN·m

Question 9

A concentrated applied couple on a beam normally produces what diagram feature?

A. A jump in the bending-moment diagram

B. A jump in axial force only

C. A zero shear everywhere

D. A change in span length

Correct Answer: A. A jump in the bending-moment diagram

Question 10

Where does a smooth local maximum bending moment commonly occur?

A. Where the shear force is zero

B. Where the axial force is zero only

C. At every roller support

D. At the point of greatest span length

Correct Answer: A. Where the shear force is zero