Determinate Beams and Internal Forces

Unit purpose

A determinate beam is an ideal setting in which to connect equilibrium with the diagrams used in flexural design and serviceability checks. What matters is not memorizing four equations but building intuition about resultant location, sign conventions, discontinuities and stiffness.

Field scenario

A crane runway often combines near-concentrated wheel forces and distributed girder weight. A support reaction can be highly sensitive to wheel location, while deflection is particularly sensitive to span and stiffness. Keep these separate as you evaluate the examples.

Topics and worked reasoning

Topic 2.1: Reactions and equivalent beam loads

For a simply supported prismatic beam of span L with downward point force P at distance a from the left support and downward uniform line load w over the full span, the support reactions are R_A=P(L-a)/L+wL/2 and R_B=Pa/L+wL/2. These follow from moments about the opposite support and vertical force equilibrium. Reactions do not depend on bending stiffness for this statically determinate model. An incorrect location of an equivalent load is a common error: the resultant of a uniform load acts at its centroid, but triangular and trapezoidal distributions have different centers of action.

Engineering practice: Work first symbolically, then substitute numerical values. Treat P in kN, w in kN/m and L in m so reactions are kN. Check that R_A+R_B=P+wL and that the moment sum is zero. If the point load moves toward one support, the near support reaction must increase; if a model reports the opposite, inspect the load position convention.

Fully evaluated study example

For span 8.0 m, load 24 kN at x=5.5 m and uniform line load 0.9 kN/m, RA=11.100 kN. At the point-force section the bending moment is M(a)=47.438 kN·m, while the shear jumps downward by 24.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Reasoning and interpretation

Where does an equivalent point resultant act for each load shape? Start by writing the physical reason, then use an equation or laboratory parameter sweep to support the claim. Compare the result against an extreme case: zero applied load, negligible rise, very long span, perfectly fixed restraint, or load at a support, whichever is physically appropriate. In a valid limiting case, the response should approach a predictable behavior. When a calculation fails an extreme-case check, investigate modeling assumptions before blaming numerical precision.

Practice tasks

1. Rework the given example with a 15% increase in the characteristic load and predict whether force, moment or displacement scales linearly. State all inputs kept fixed.
2. Investigate this potential failure of the model: Where does an equivalent point resultant act for each load shape? Document the assumption at risk and describe a more complete model.
3. Draw the free-body diagram or curvature/influence diagram appropriate to reactions and equivalent beam loads. Label axes, internal forces, signs, physical units and boundary conditions.
4. Use a unit or equilibrium residual to check your solution. Describe an error that would pass the numerical residual check but still lead to an incorrect engineering interpretation.

Applied investigation

Choose a structural example relevant to reactions and equivalent beam loads. List the information required from a drawing, material schedule or site measurement. Separate what can be concluded directly from equilibrium from what depends on stiffness, geometry, or an unverified boundary condition. Run two different reasonable parameter settings and write a short explanation of why the predicted response changes. Avoid treating either result as a certified design action.

Topic 2.2: Shear force and bending moment fields

Using upward forces positive on a left segment and sagging moments positive, V(x)=R_A-wx-PH(x-a), with H the step function, and M(x)=R_Ax-wx²/2-P max(0,x-a) for a beam with the above loads. A point force produces a jump in shear and a corner in the moment curve; a uniform load creates linear shear and parabolic moment. An ideal concentrated couple would create a jump in bending moment but not in shear. The derivative identities dM/dx=V and dV/dx=-w are more powerful than memorizing diagram shapes, but they apply away from singular applied actions.

Engineering practice: For a candidate maximum bending moment, locate points where shear changes sign, where a point-force jump spans zero, and at any geometric boundaries. Do not assume the span midpoint is always critical. Check M(0)=M(L)=0 for an ideal simple span without applied end couples. Diagrams should carry signs, units and discontinuity conventions, not just a positive envelope.

Fully evaluated study example

For span 10.0 m, load 45 kN at x=5.0 m and uniform line load 2.4 kN/m, RA=34.500 kN. At the point-force section the bending moment is M(a)=142.500 kN·m, while the shear jumps downward by 45.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Reasoning and interpretation

Does shear jump at a concentrated force while moment stays continuous? Start by writing the physical reason, then use an equation or laboratory parameter sweep to support the claim. Compare the result against an extreme case: zero applied load, negligible rise, very long span, perfectly fixed restraint, or load at a support, whichever is physically appropriate. In a valid limiting case, the response should approach a predictable behavior. When a calculation fails an extreme-case check, investigate modeling assumptions before blaming numerical precision.

Practice tasks

1. Rework the given example with a 15% increase in the characteristic load and predict whether force, moment or displacement scales linearly. State all inputs kept fixed.
2. Investigate this potential failure of the model: Does shear jump at a concentrated force while moment stays continuous? Document the assumption at risk and describe a more complete model.
3. Draw the free-body diagram or curvature/influence diagram appropriate to shear force and bending moment fields. Label axes, internal forces, signs, physical units and boundary conditions.
4. Use a unit or equilibrium residual to check your solution. Describe an error that would pass the numerical residual check but still lead to an incorrect engineering interpretation.

Applied investigation

Choose a structural example relevant to shear force and bending moment fields. List the information required from a drawing, material schedule or site measurement. Separate what can be concluded directly from equilibrium from what depends on stiffness, geometry, or an unverified boundary condition. Run two different reasonable parameter settings and write a short explanation of why the predicted response changes. Avoid treating either result as a certified design action.

Topic 2.3: Beam deflection and stiffness

Under Euler–Bernoulli assumptions the moment–curvature relation is approximately EI v double-prime equal to the signed bending moment. E is Young modulus, I the second moment of cross-sectional area and v the transverse displacement with the chosen sign convention. For a simply supported beam with uniform load w, the maximum deflection at midspan is 5wL⁴/(384EI). For a central point load P, the midspan deflection is PL³/(48EI). These results rely on constant EI, linear elasticity, small rotations and ideal boundary supports. Because displacement depends on the fourth power of span for a uniform load, modest increases in span can markedly increase serviceability demand.

Engineering practice: Perform a dimensional check using consistent N–m units before computing mm. Use a second method to validate one section, such as integrating the moment function with support displacement conditions. Explain why a large numerical deflection relative to the span invalidates the small-deflection assumptions used to derive the formula.

Fully evaluated study example

For span 7.5 m, load 38 kN at x=4.5 m and uniform line load 3.8 kN/m, RA=29.450 kN. At the point-force section the bending moment is M(a)=94.050 kN·m, while the shear jumps downward by 38.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Reasoning and interpretation

Which boundary condition fixes each integration constant? Start by writing the physical reason, then use an equation or laboratory parameter sweep to support the claim. Compare the result against an extreme case: zero applied load, negligible rise, very long span, perfectly fixed restraint, or load at a support, whichever is physically appropriate. In a valid limiting case, the response should approach a predictable behavior. When a calculation fails an extreme-case check, investigate modeling assumptions before blaming numerical precision.

Practice tasks

1. Rework the given example with a 15% increase in the characteristic load and predict whether force, moment or displacement scales linearly. State all inputs kept fixed.
2. Investigate this potential failure of the model: Which boundary condition fixes each integration constant? Document the assumption at risk and describe a more complete model.
3. Draw the free-body diagram or curvature/influence diagram appropriate to beam deflection and stiffness. Label axes, internal forces, signs, physical units and boundary conditions.
4. Use a unit or equilibrium residual to check your solution. Describe an error that would pass the numerical residual check but still lead to an incorrect engineering interpretation.

Applied investigation

Choose a structural example relevant to beam deflection and stiffness. List the information required from a drawing, material schedule or site measurement. Separate what can be concluded directly from equilibrium from what depends on stiffness, geometry, or an unverified boundary condition. Run two different reasonable parameter settings and write a short explanation of why the predicted response changes. Avoid treating either result as a certified design action.

Topic 2.4: Section actions and load sensitivity

A beam does not experience a single bending moment; it contains a field of axial force, shear and bending moment along its length. A design section may be governed by a localized shear peak while a different section governs flexure. Reactions and moments of a determinate beam vary linearly with applied load magnitude, but stiffness affects displacement and stress distribution among connected indeterminate members. A stiffness change that leaves equilibrium reactions untouched can substantially alter deflection. This distinction is central to evaluating whether strengthening by a larger section is effective.

Engineering practice: Compare the sensitivity of maximum shear, maximum moment and midspan deflection to doubling the span while keeping the line-load intensity fixed. For a uniform simple beam, reactions scale with L, moment with L squared and displacement with L to the fourth power if EI remains fixed. Record which structural response should be monitored during any span optimization.

Fully evaluated study example

For span 5.5 m, load 19 kN at x=3.5 m and uniform line load 2.1 kN/m, RA=12.684 kN. At the point-force section the bending moment is M(a)=31.532 kN·m, while the shear jumps downward by 19.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Reasoning and interpretation

Why might a larger section improve deflection but not beam reactions? Start by writing the physical reason, then use an equation or laboratory parameter sweep to support the claim. Compare the result against an extreme case: zero applied load, negligible rise, very long span, perfectly fixed restraint, or load at a support, whichever is physically appropriate. In a valid limiting case, the response should approach a predictable behavior. When a calculation fails an extreme-case check, investigate modeling assumptions before blaming numerical precision.

Practice tasks

1. Rework the given example with a 15% increase in the characteristic load and predict whether force, moment or displacement scales linearly. State all inputs kept fixed.
2. Investigate this potential failure of the model: Why might a larger section improve deflection but not beam reactions? Document the assumption at risk and describe a more complete model.
3. Draw the free-body diagram or curvature/influence diagram appropriate to section actions and load sensitivity. Label axes, internal forces, signs, physical units and boundary conditions.
4. Use a unit or equilibrium residual to check your solution. Describe an error that would pass the numerical residual check but still lead to an incorrect engineering interpretation.

Applied investigation

Choose a structural example relevant to section actions and load sensitivity. List the information required from a drawing, material schedule or site measurement. Separate what can be concluded directly from equilibrium from what depends on stiffness, geometry, or an unverified boundary condition. Run two different reasonable parameter settings and write a short explanation of why the predicted response changes. Avoid treating either result as a certified design action.

Deep dive and misconception repair

Diagram clinic: discontinuities and extrema

Consider a simply supported beam with a point force positioned away from midspan and a full-span distributed load. A useful hand solution begins with reactions and then writes piecewise expressions for shear and moment. The shear to the left of the point action differs from the shear to its right by exactly the downward point-force magnitude. The bending moment is continuous through that location if no concentrated couple is applied. If a numerical chart instead shows a moment jump at the point-force coordinate, its implementation likely conflates force with couple.

The maximum positive moment need not occur where the largest force is applied. In regions where shear is positive, moment grows with x. Where shear is negative, moment decreases with x. A point-load jump in shear may cross zero abruptly, creating a local moment maximum at the load coordinate. Otherwise one can solve V(x)=0 in each smooth region and compare the resulting candidate moments. This logic generalizes to piecewise load functions and avoids blind reliance on sampled graphs.

Draw the distributed load and the beam's resultant separately. When computing reactions, a uniform line load can be replaced by a single resultant at the centroid, because global equilibrium depends on the resultant and its moment. When calculating internal shear and moment as functions of x, however, the replacement cannot be used as if it were a point force acting at all partial beam cuts. A cut at x encloses only the portion of distributed load to its left; the correct force is wx, not the total wL.

For a beam with a pure end couple, the bending-moment boundary conditions differ from an unloaded pin. The internal diagram can include a nonzero end moment consistent with the applied couple, despite a pin support that cannot resist a couple. The statement 'moment at a pin equals zero' applies to the reaction transmitted by a pin, not to an arbitrary applied couple at the same coordinate. This distinction often prevents an apparently contradictory free-body diagram.

Sensitivity study: two independent scalings

If a beam carries only a constant uniform load w and is geometrically similar as its span changes, the total load increases with L. The support reactions each equal wL/2. The maximum sagging moment equals wL²/8, located at midspan. With constant E and I, the maximum deflection equals 5wL⁴/(384EI). Doubling span while keeping w, E and I fixed multiplies reaction by two, bending moment by four and deflection by sixteen. This is why checking only strength can overlook serviceability for a long slender span.

If instead the total uniform load W is held constant while changing span, the line intensity is w=W/L. Reactions remain W/2, maximum moment scales as WL/8 and deflection scales as 5WL³/(384EI). The two cases answer different engineering questions; mixing their scaling conclusions is a common conceptual error. Always state whether total action or distributed intensity is held fixed.

The simple-beam deflection formula is derived under small curvature, constant stiffness, linear elastic material and supports that prevent vertical displacement but not rotation. In practice, shear flexibility can matter for deep beams, supports can settle, and concrete stiffness may vary after cracking. The correct use of a closed form is to understand the mechanism and assess whether its assumptions are adequate, not to disguise those assumptions behind several decimal places.

Multi-case calculations and validation

Case 2.1: Reaction shift

For span 6.0 m, load 28 kN at x=2.0 m and uniform line load 3.1 kN/m, RA=27.967 kN. At the point-force section the bending moment is M(a)=49.733 kN·m, while the shear jumps downward by 28.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.2: Longer girder

For span 9.0 m, load 34 kN at x=6.0 m and uniform line load 2.0 kN/m, RA=20.333 kN. At the point-force section the bending moment is M(a)=86.000 kN·m, while the shear jumps downward by 34.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.3: Footbridge loading

For span 12.0 m, load 41 kN at x=4.0 m and uniform line load 4.8 kN/m, RA=56.133 kN. At the point-force section the bending moment is M(a)=186.133 kN·m, while the shear jumps downward by 41.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.4: Short lintel

For span 4.5 m, load 15 kN at x=1.5 m and uniform line load 1.2 kN/m, RA=12.700 kN. At the point-force section the bending moment is M(a)=17.700 kN·m, while the shear jumps downward by 15.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.5: Beam with equipment

For span 8.0 m, load 24 kN at x=5.5 m and uniform line load 0.9 kN/m, RA=11.100 kN. At the point-force section the bending moment is M(a)=47.438 kN·m, while the shear jumps downward by 24.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.6: Symmetric midspan

For span 10.0 m, load 45 kN at x=5.0 m and uniform line load 2.4 kN/m, RA=34.500 kN. At the point-force section the bending moment is M(a)=142.500 kN·m, while the shear jumps downward by 45.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.7: Warehouse girder

For span 7.5 m, load 38 kN at x=4.5 m and uniform line load 3.8 kN/m, RA=29.450 kN. At the point-force section the bending moment is M(a)=94.050 kN·m, while the shear jumps downward by 38.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Case 2.8: Roof beam

For span 5.5 m, load 19 kN at x=3.5 m and uniform line load 2.1 kN/m, RA=12.684 kN. At the point-force section the bending moment is M(a)=31.532 kN·m, while the shear jumps downward by 19.0 kN. At x=0 the bending moment is zero; at x=L it returns to zero to within floating-point rounding. The maximum positive bending moment can occur where shear passes through zero, not necessarily at x=a. A full scan must examine both sides of that discontinuity.

Verification challenge: Check the key result above against a limiting case in which one action is reduced to zero, or one geometric parameter is increased. For determinate beams and internal forces, identify the response that changes most and state the governing physical mechanism. Do not use this case as a final capacity evaluation; it omits detailed component resistance and code-required factors.

Interactive experiments

Use these standalone HTML laboratories. They load without a web connection and export response curves in CSV format. Each lab shows a calculation formula and limitations.

• [Simulation 01: Beam reactions and load balance](../labs/01-beam-reactions.html)
• [Simulation 02: Shear and bending-moment fields](../labs/02-shear-moment.html)
• [Simulation 03: Elastic beam deflection laboratory](../labs/03-beam-deflection.html)

Unit assessment

1. Explain the key idealization governing determinate beams and internal forces, and describe one condition that would invalidate it.
2. Explain what mathematical or physical equilibrium, compatibility, or limiting-case checks are appropriate for this unit.
3. Compare the results of two parameter settings from a linked lab. Name the inputs, calculate one numerical difference and interpret its sign.
4. Provide one real-world detail that the model does not represent and specify how an engineer should investigate it.

Project brief

Prepare a two-page calculation note using the methods of determinate beams and internal forces. The note must include a problem definition, a labeled free-body diagram, a parameter and unit table, equations, at least two worked numerical results, an independently evaluated check, a figure or CSV trace from a linked simulation, and a discussion of three limitations. The note is not a structural design certification.