Structural Analysis of a Simply Supported Beam: Reactions, Shear Force, Bending Moment and Deflection

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1. Introduction

The simply supported beam is one of the most widely encountered members in structural and mechanical engineering, appearing in floor joists, bridge girders, crane runways and machine frames. Understanding how such a member responds to transverse loading is fundamental to safe design: the engineer must be able to establish the support reactions, trace the internal shear force and bending moment along the span, identify the location and magnitude of the maximum bending moment, verify that the resulting bending stress is acceptable, and confirm that deflection remains within serviceability limits. This assignment works through each of these steps for a defined beam scenario, applying the standard formulae of elementary beam theory and stating the assumptions on which that theory rests.

The analysis presented here is a worked example. A specific beam geometry, loading and cross-section have been selected so that every quantity can be computed explicitly and checked. The methods, however, are general and transfer directly to other statically determinate beams.

2. Assumptions and theoretical basis

The results that follow rest on the assumptions of engineers’ theory of bending, sometimes called the Euler–Bernoulli beam theory (Gere and Goodno, 2013):

1. Linear elastic material. The beam material obeys Hooke’s law throughout, so stress is proportional to strain and the modulus of elasticity E is constant. No yielding occurs. 2. Small deflections. Deflections are small relative to the span, so the curvature may be approximated by the second derivative of the deflection and the principle of superposition applies (Hibbeler, 2018). 3. Plane sections remain plane. Cross-sections that are plane and normal to the neutral axis before bending remain plane and normal after bending, giving a linear distribution of bending strain through the depth. 4. Prismatic beam. The cross-section and material are uniform along the length; E and the second moment of area I are constant. 5. Loading in the plane of symmetry. Loads act in the vertical plane containing the centroidal axis, so the beam bends without twisting. 6. Self-weight of the beam itself is neglected, as the applied loading dominates and the section is comparatively light.

Under these assumptions the beam is statically determinate: the two unknown reactions can be found from the two available equations of planar static equilibrium.

3. Beam scenario (worked example)

A horizontal steel beam of span L = 6.0 m is simply supported, with a pin support at end A and a roller support at end B. It carries two loads acting together:

• a concentrated (point) load P = 20 kN applied at a distance a = 2.0 m from support A (so that b = 4.0 m from support B), and
• a uniformly distributed load (UDL) of intensity w = 5 kN/m acting over the whole span.

The chosen cross-section is a rolled universal beam, the 305 × 165 × 40 UB, with second moment of area I = 8 500 cm⁴ = 8.50 × 10⁻⁵ m⁴ about the axis of bending (noting that 1 cm⁴ = 10⁻⁸ m⁴) and overall depth d = 303.4 mm. The material is structural steel with modulus of elasticity E = 200 GPa (Hibbeler, 2018).

The objective is to determine the reactions, the shear force and bending moment diagrams, the maximum bending moment and corresponding bending stress, and the maximum deflection.

4. Support reactions

The total applied load consists of the point load and the resultant of the UDL. The UDL resultant is

$$W = wL = 5 \times 6.0 = 30\ \text{kN},$$

acting at the mid-span, 3.0 m from either support.

Taking moments about A (anticlockwise positive) and setting the sum to zero:

$$\sum M_A = 0: \quad R_B \times L – P \times a – W \times \frac{L}{2} = 0$$

$$R_B = \frac{P a + W (L/2)}{L} = \frac{(20)(2.0) + (30)(3.0)}{6.0} = \frac{40 + 90}{6.0} = 21.667\ \text{kN}.$$

Vertical equilibrium then gives the reaction at A:

$$\sum F_y = 0: \quad R_A + R_B – P – W = 0$$

$$R_A = P + W – R_B = 20 + 30 – 21.667 = 28.333\ \text{kN}.$$

As a check, taking moments about B independently yields the same value, $R_A = [(20)(4.0) + (30)(3.0)]/6.0 = 28.333$ kN, and the two reactions sum to the total load of 50 kN, confirming equilibrium.

Reactions: R_A = 28.33 kN, R_B = 21.67 kN (both acting upward).

5. Shear force diagram

The shear force V at a section is the algebraic sum of the vertical forces to the left of that section, taking upward forces as positive. Moving from A to B:

• Just to the right of A (x = 0⁺): the only force to the left is R_A, so V = +28.33 kN.
• Approaching the point load (x = 2.0⁻ m): the UDL removes 5 kN/m over 2.0 m, i.e. 10 kN, so V = 28.33 − 10 = +18.33 kN. Between A and the load the shear falls linearly.
• At the point load, the shear drops by P = 20 kN, giving V = 18.33 − 20 = −1.67 kN just to the right of the load (x = 2.0⁺ m).
• Just to the left of B (x = 6.0⁻ m): V = 28.33 − wL − P = 28.33 − 30 − 20 = −21.67 kN, which equals −R_B, as it must.

The shear force therefore starts at +28.33 kN, slopes down linearly under the UDL to +18.33 kN at the load, steps down abruptly by 20 kN to −1.67 kN, and continues sloping down to −21.67 kN at B, where the reaction R_B restores it to zero. The diagram crosses zero at the point load, because the shear changes sign from positive to negative there; this crossing locates the maximum bending moment (Hibbeler, 2018).

6. Bending moment diagram

The bending moment M at a distance x from A (for the region up to the load, x ≤ a) is

$$M(x) = R_A x – \frac{w x^2}{2},$$

and beyond the load (x > a) an extra term −P(x − a) is subtracted. Evaluating at key sections (all in kN·m):

Because the beam is simply supported, the moment is zero at both ends. It rises as a curve (parabolic under the UDL, with the point load introducing a change of slope) to a peak directly beneath the concentrated load, then falls back to zero at B. The bending moment diagram is sagging (positive) throughout, so the beam hogs nowhere and tension develops on the underside along the whole span.

7. Maximum bending moment

The maximum bending moment occurs where the shear force passes through zero. Here that point coincides with the position of the concentrated load, x = a = 2.0 m, because the shear changes sign at the load itself. Substituting x = 2.0 m:

$$M_{max} = R_A a – \frac{w a^2}{2} = (28.333)(2.0) – \frac{(5)(2.0)^2}{2} = 56.667 – 10.0 = 46.667\ \text{kN·m}.$$

Maximum bending moment M_max = 46.67 kN·m at x = 2.0 m from A.

8. Maximum bending stress

The bending (flexural) stress at the extreme fibre is given by the flexure formula (Gere and Goodno, 2013):

$$\sigma_{max} = \frac{M_{max}\, c}{I} = \frac{M_{max}}{Z},$$

where c is the distance from the neutral axis to the extreme fibre and Z = I/c is the elastic section modulus. For the symmetric 305 × 165 × 40 UB, c = d/2 = 303.4/2 = 151.7 mm = 0.1517 m, giving

$$Z = \frac{I}{c} = \frac{8.50 \times 10^{-5}}{0.1517} = 5.60 \times 10^{-4}\ \text{m}^3 = 560\ \text{cm}^3.$$

The maximum bending stress is then

$$\sigma_{max} = \frac{M_{max} c}{I} = \frac{(46\,667)(0.1517)}{8.50 \times 10^{-5}} = 83.3 \times 10^{6}\ \text{Pa} = 83.3\ \text{MPa}.$$

For grade S275 structural steel the design yield strength is 275 MPa (BSI, 2005). The working bending stress of 83.3 MPa therefore represents a factor of safety of approximately 275/83.3 ≈ 3.3 against yielding in bending, which is comfortable and indicates the section is not over-stressed.

9. Maximum deflection

Because the material is linear elastic and deflections are small, the total deflection is obtained by superposition of the deflections due to the UDL and the point load acting separately (Gere and Goodno, 2013).

UDL contribution. For a simply supported beam under a full-span UDL the maximum deflection occurs at mid-span and is

$$\delta_{UDL} = \frac{5 w L^4}{384\, E I} = \frac{5 (5\,000)(6.0)^4}{384 (200 \times 10^{9})(8.50 \times 10^{-5})} = 4.96 \times 10^{-3}\ \text{m} = 4.96\ \text{mm}.$$

Point load contribution. For a concentrated load P at distance a from A (and b from B), the deflection at any section is given by the standard equations (Hibbeler, 2018). Evaluating them along the span, the point-load deflection at mid-span is 4.51 mm and its own maximum is 4.55 mm, occurring close to mid-span.

Total deflection. Superposing the two effects at each section, the maximum total deflection is

$$\delta_{max} = 9.49\ \text{mm, occurring at } x \approx 2.87\ \text{m from A},$$

which is only marginally larger than the mid-span value of 9.47 mm; for practical purposes the maximum deflection may be taken as approximately 9.5 mm at mid-span.

Serviceability check. A commonly applied limit for beams carrying general loads is span/250 (BSI, 2005). Here

$$\frac{L}{250} = \frac{6\,000}{250} = 24.0\ \text{mm}.$$

The computed deflection of 9.5 mm is well within this limit (and also within the stricter span/360 = 16.7 mm often applied to brittle finishes), so the beam satisfies the deflection serviceability requirement with a good margin.

10. Discussion

The analysis shows the beam is governed comfortably by strength and stiffness alike: the maximum bending stress of 83.3 MPa sits at roughly one-third of the S275 yield strength, and the peak deflection of 9.5 mm is under half the span/250 serviceability limit. The maximum bending moment falls at the concentrated load rather than at mid-span, a direct consequence of the shear force changing sign there; this reinforces the general rule that the bending moment is stationary wherever the shear is zero, which for a point load means at the load itself.

Several idealisations should be borne in mind when interpreting these figures. Neglecting self-weight is reasonable for a light rolled section but would need revisiting for a heavy or long-span member. The assumption of purely elastic behaviour is valid here because the stresses are modest, but at higher load a plastic analysis, giving a larger reserve of strength, would be more representative. The small-deflection assumption is clearly justified, since the deflection is roughly 1/630 of the span. Finally, ideal pin and roller supports have been assumed; real connections may offer partial rotational restraint, which would reduce mid-span moment and deflection and introduce end moments not captured by the simply supported model.

11. Conclusion

For the defined 6 m simply supported steel beam carrying a 20 kN point load at 2 m together with a 5 kN/m UDL, the support reactions are R_A = 28.33 kN and R_B = 21.67 kN. The shear force varies from +28.33 kN at A to −21.67 kN at B, changing sign at the point load, where the bending moment reaches its maximum of 46.67 kN·m. The corresponding maximum bending stress in the 305 × 165 × 40 UB section is 83.3 MPa, a factor of about 3.3 below yield, and the maximum deflection is approximately 9.5 mm, well inside the span/250 serviceability limit of 24 mm. The beam and section chosen are therefore adequate in both strength and stiffness for the given loading. The worked example demonstrates the complete elementary procedure — equilibrium for reactions, section analysis for internal forces, the flexure formula for stress, and superposition for deflection — that underpins the design of statically determinate beams.

References

BSI (2005) BS EN 1993-1-1:2005 Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings. London: British Standards Institution.

Gere, J.M. and Goodno, B.J. (2013) Mechanics of Materials. 8th edn. Stamford, CT: Cengage Learning.

Hibbeler, R.C. (2018) Mechanics of Materials. 10th edn. Harlow: Pearson Education.

The Steel Construction Institute (2019) Steel Building Design: Design Data (SCI P363). Ascot: SCI.