---
title: "Statically determinate/indeterminate; beam SFD and BMD"
subject: "Applied Mechanics"
chapter: 6
key: "semester_1/applied_mechanics:determinate-indeterminate-beam-sfd-bmd-triangular"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/determinate-indeterminate-beam-sfd-bmd-triangular
site: BCT Engineering
---

# Statically determinate/indeterminate; beam SFD and BMD

Applied Mechanics · Chapter 6: Analysis of Beams and Frames

- Asked in: 2069 Asar, 4+10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/determinate-indeterminate-beam-sfd-bmd-triangular
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-6

## Question

What do you understand by statically determinate and indeterminate structure? Explain with an example for each. Write the equations for shear force and bending moment for the beam shown in figure below. Plot the variation of the shear force and bending moment in the beam. Also indicate the salient features associated with the shear force and bending moment. [Figure: simply supported beam, roller at left and pin at right; triangular load from 0 to W = 15 kN/m over the first 1.8 m + 3.6 m; 65 kN point load; segments 1.8 m, 3.6 m, 1.2 m, 1.2 m.]

## Short answer

A statically determinate structure has unknowns equal to equilibrium equations (e.g., simply supported beam); an indeterminate structure has more unknowns, requiring compatibility equations (e.g., propped cantilever). For the beam, reactions are RA = 41.35 kN↑, RB = 64.15 kN↑. Maximum bending moment is 76.98 kN·m at x = 6.6 m under the 65 kN load; a contraflexure point occurs at x ≈ 1.87 m.

## Answer

#### Statically determinate and indeterminate structures
A structure is **statically determinate** if all support reactions and internal forces can be found from the equations of static equilibrium alone (ΣFx = 0, ΣFy = 0, ΣM = 0 in 2D). It is **statically indeterminate** when the unknowns exceed the available equations, so extra equations from deformation (compatibility) are needed.

| | Determinate | Indeterminate |
|---|---|---|
| Unknown reactions (beam) | r = 3 (or r = 3 + number of hinge conditions) | r > 3 + hinge conditions |
| Method | Equilibrium only | Equilibrium + compatibility (slope-deflection, moment distribution, flexibility) |
| Effect of settlement, temperature | No stresses | Stresses develop |
| Example | Simply supported beam (pin + roller, r = 3) | Propped cantilever or continuous beam (r = 4 or more) |

Degree of indeterminacy of a beam: D = r − 3 − (number of internal hinges). A propped cantilever has r = 3 + 1 = 4, so D = 1.

#### Reading of the figure
Origin at the free left end O; roller at A (x = 1.8 m), pin at B (x = 7.8 m). The triangular load rises from 0 at O to w = 15 kN/m at x = 5.4 m (intensity 2.778x kN/m); a 65 kN point load acts at x = 6.6 m.
- Total triangular load = ½ × 15 × 5.4 = 40.5 kN, acting at 2/3 × 5.4 = 3.6 m from O.

```text
   w=15 kN/m               65 kN
  /||||||||||||||           ↓
 O/_______________________________
 ^  A(roller)                      B(pin)
 |<-1.8->|<--3.6-->|<1.2>|<1.2>|
```

#### Reactions
- ΣM_B = 0: R_A(7.8 − 1.8) = 40.5(7.8 − 3.6) + 65(1.2), so 6R_A = 170.1 + 78 = 248.1 and **R_A = 41.35 kN ↑**.
- ΣFy = 0: R_B = 40.5 + 65 − 41.35 = **64.15 kN ↑**. (r = 3, so the beam is determinate.)

#### Equations of V and M (x measured from O)
Load intensity w(x) = (15/5.4)x = 2.778x kN/m, load up to x = ½·2.778x·x = 1.389x².

| Range | Shear force V (kN) | Bending moment M (kN·m) |
|---|---|---|
| 0 ≤ x ≤ 1.8 | −1.389x² | −0.463x³ |
| 1.8 ≤ x ≤ 5.4 | 41.35 − 1.389x² | 41.35(x − 1.8) − 0.463x³ |
| 5.4 ≤ x ≤ 6.6 | 0.85 | 0.85x + 71.37 |
| 6.6 ≤ x ≤ 7.8 | −64.15 | 64.15(7.8 − x) |

(Sagging moment and the usual sign convention: upward force on the left of the section gives positive shear.)

#### Ordinates
| x (m) | V (kN) | M (kN·m) |
|---|---|---|
| 0 | 0 | 0 |
| 1.8⁻ (left of A) | −4.50 | −2.70 |
| 1.8⁺ (right of A) | +36.85 | −2.70 |
| 3.6 | 23.35 | 41.35(1.8) − 0.463(3.6)³ = 52.83 |
| 5.4 | 0.85 | 75.96 |
| 6.6⁻ | 0.85 | 76.98 |
| 6.6⁺ | −64.15 | 76.98 |
| 7.8 | −64.15 | 0 |

#### Diagrams and salient features
```text
 SFD (kN):  +36.85 just right of A
            decreases (parabola) to +0.85 at x = 5.4
            stays +0.85 to x = 6.6, jumps to −64.15
            constant −64.15 to B

 BMD (kN·m): 0 at O, −2.70 at A (hogging),
            0 at x = 1.87, +75.96 at x = 5.4,
            +76.98 at x = 6.6 (maximum), 0 at B
```
- Shear changes abruptly by +41.35 kN at the roller A, and by −65 kN at the point load (from +0.85 to −64.15 kN).
- The shear is 0.85 kN, almost zero, between x = 5.4 m and 6.6 m, so the bending moment is nearly flat there: the **maximum moment is 76.98 kN·m at x = 6.6 m** (under the 65 kN load), where the shear changes sign.
- A small hogging moment of 2.70 kN·m occurs over the support A (cantilever overhang); the moment changes sign at x ≈ 1.87 m (point of contraflexure, from 41.35(x − 1.8) = 0.463x³).
- Shear is a parabola over the triangular load (V falls by 1.389x²); the moment is a cubic curve there, a straight line under no load.
- The moment is zero at the pin B (end of the simply supported span).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A structure is statically determinate when:
   A. It has no supports
   B. It can be analysed using compatibility of deformation
   C. All reactions and internal forces can be found from equilibrium equations alone
   D. It has more reactions than equations
   Answer: C. Determinate structures need only ΣFx = 0, ΣFy = 0, ΣM = 0. Indeterminate ones need extra compatibility equations. (Key point: Determinate structure)

2. A propped cantilever has r = 4. What is its degree of indeterminacy?
   A. 0
   B. 1
   C. 2
   D. 3
   Answer: B. D = r - 3 - (hinges) = 4 - 3 = 1, so it is indeterminate to the first degree. (Key point: Degree of indeterminacy)

3. Where is the maximum bending moment (76.98 kN·m) in the beam?
   A. Under the 65 kN load at x = 6.6 m, where shear changes sign
   B. At the roller A
   C. At the pin B
   D. At the end of the triangular load, x = 5.4 m
   Answer: A. Moment is maximum where shear changes sign. Shear drops from +0.85 to -64.15 kN under the point load, so the maximum is at x = 6.6 m. (Key point: Maximum bending moment)

4. Over the triangular load, what shape do the shear and bending moment curves have?
   A. Shear linear, moment parabolic
   B. Shear parabolic, moment cubic
   C. Both linear
   D. Shear constant, moment linear
   Answer: B. Load w = 2.778x gives V = -1.389x^2 (parabola) and M = -0.463x^3 (cubic). Without load, V is constant and M linear. (Key point: Shape of V and M diagrams)

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