---
title: "Compatibility condition; portal frame temperature variation"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:compatibility-condition-portal-temperature"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/compatibility-condition-portal-temperature
site: BCT Engineering
---

# Compatibility condition; portal frame temperature variation

Theory of Structures II · Chapter 2: Force method

- Asked in: 2071 Shrawan, 15 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/compatibility-condition-portal-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Explain the physical meaning of the compatibility condition and derive the equation for it. A portal frame with hinged supports is subjected to a temperature variation as shown in figure below. Determine flexibility coefficients and calculate the redundant force with the help of the compatibility equation. Take $\alpha = 11\times10^{-6}/°C$, $E = 5000\sqrt{f_{ck}}$, $f_{ck} = 20$ MPa and constant flexural rigidity. [Figure: portal frame ABCD with hinged supports at A and D, height 6 m, span 3 m; temperature $t_1 = 20°C$ on the outer face and $t_2 = 10°C$ on the inner face of the beam; member cross-section 0.3 m x 0.6 m.]

## Short answer

Compatibility requires the redundant to make the displacement at the release equal the actual support displacement. For the portal, f11=2.087e-3 m/kN, temperature displacement Δ1t=-2.805e-3 m, so H=-Δ1t/f11=1.344 kN outward, with E=22360.7 MPa and EI=120748 kNm^2.

## Answer

#### Physical meaning of the compatibility condition
A statically indeterminate structure has more supports or members than are needed for equilibrium. When the redundants are removed, the primary structure deforms freely, and its displacements at the released points generally contradict the real conditions of the structure (for example, a fixed or hinged support cannot move). The **compatibility condition** states that the redundants must be of such a size that the displacements of the real structure at the releases agree with the real support conditions (usually zero, or equal to the settlement). It ensures that the deformed shape is continuous and fits the supports.

#### Derivation (one redundant $X_1$)
Let the primary structure carry the loads and the unknown $X_1$. By superposition, the displacement at the release point in the direction of $X_1$ is

$$
\Delta_1 = \Delta_{10} + f_{11}X_1
$$

where $\Delta_{10}$ = displacement of the primary structure due to the loads (and temperature, settlement of other supports, lack of fit) and $f_{11}$ = displacement due to a unit $X_1$. Compatibility requires $\Delta_1$ to equal the actual displacement $\delta_1$ of the support (zero for a rigid support):

$$
\Delta_{10} + f_{11}X_1 = \delta_1 \quad\Rightarrow\quad X_1 = \frac{\delta_1 - \Delta_{10}}{f_{11}}
$$

For $n$ redundants: $\{\Delta_0\} + [F]\{X\} = \{\delta\}$, with $f_{ij} = \int \dfrac{m_im_j}{EI}dx$ and $\Delta_{i0} = \int \dfrac{M_0m_i}{EI}dx$. For a temperature change, the load term is

$$
\Delta_{it} = \int m_i\,\frac{\alpha\,(t_1 - t_2)}{d}\,dx + \int n_i\,\alpha\,t_m\,dx,\qquad t_m = \frac{t_1+t_2}{2}
$$

where $d$ is the depth, $m_i$ the unit BM and $n_i$ the unit axial force.

#### Portal frame with temperature variation
Reading of the figure: hinged portal ABCD, height $h = 6$ m, span $L = 3$ m, EI constant. The beam is warmer on the outer (top) face by $t_1 = 20^\circ$C against $t_2 = 10^\circ$C on the inner face. Section $0.3\times0.6$ m ($d = 0.6$ m). Temperatures are changes from the original state.

$D_s = 4 - 3 = 1$. Release D horizontally and take the thrust $H$ at D (positive outward) as the redundant.

```text
   B ____________ C   t1 = 20 (outer)
   |              |   t2 = 10 (inner)
   |              |
   | 6 m          | 6 m
   |              |
  A o            o D <- H
   |<--- 3 m --->|
```

#### Elastic constants
$$
E = 5000\sqrt{f_{ck}} = 5000\sqrt{20} = 22360.7\ \text{MPa} = 22360680\ \text{kN/m}^2
$$

$$
I = \frac{0.3\times0.6^3}{12} = 0.0054\ \text{m}^4,\qquad EI = 120748\ \text{kNm}^2
$$

#### Flexibility coefficient
A unit outward force at D gives $m = y$ in each column (from the hinge) and $m = 6$ in the beam:

$$
f_{11} = \frac{1}{EI}\left[2\cdot\frac{6^3}{3} + 6^2(3)\right] = \frac{144 + 108}{EI} = \frac{252}{EI} = 2.087\times10^{-3}\ \text{m/kN}
$$

#### Displacement of D due to temperature (primary structure)
Mean temperature of the beam $t_m = (20+10)/2 = 15^\circ$C; temperature difference $t_1 - t_2 = 10^\circ$C.

- Axial expansion of the beam (moves D outward): $\alpha\,t_m\,L = 11\times10^{-6}\times15\times3 = 4.950\times10^{-4}$ m
- Curvature $\kappa = \dfrac{\alpha(t_1-t_2)}{d} = \dfrac{11\times10^{-6}\times10}{0.6} = 1.833\times10^{-4}$ per m. The beam bends concave downward (hot face on top), which moves D inward: $-\,\kappa\,(h)(L) = -1.833\times10^{-4}\times6\times3 = -3.300\times10^{-3}$ m

$$
\Delta_{1t} = 4.950\times10^{-4} + (-3.300\times10^{-3}) = -2.805\times10^{-3}\ \text{m}\ (\text{inward})
$$

#### Compatibility ($\delta_1 = 0$)
$$
\Delta_{1t} + f_{11}H = 0 \ \Rightarrow\ H = \frac{2.805\times10^{-3}}{2.087\times10^{-3}} = 1.344\ \text{kN}
$$

The support reaction at D is **1.344 kN acting outward** (away from the frame) and an equal and opposite reaction acts at A. The vertical reactions are zero.

#### Bending moments
$M_B = M_C = H\,h = 1.344\times6 = 8.06$ kNm, with tension on the inside (sagging in the beam). The columns have a linear BM from 0 at the hinge to 8.06 kNm at the joint.

**Answer: flexibility coefficient $f_{11} = 2.087\times10^{-3}$ m/kN, thermal displacement $\Delta_{1t} = -2.805\times10^{-3}$ m, redundant $H = 1.344$ kN (outward at each hinge), corner moment $8.06$ kNm.**

## Check yourself

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

1. What does the compatibility condition ensure?
   A. Redundants are sized so displacements at the releases match the real support conditions
   B. The structure satisfies moment equilibrium at every joint
   C. Material stresses stay within the elastic limit
   D. The primary structure is statically determinate
   Answer: A. The primary structure deforms freely at the releases; the redundants must restore the real displacement (zero or settlement). (Key point: Meaning of compatibility)

2. For one redundant, which equation expresses compatibility with actual support displacement delta1?
   A. f11 X1 = Delta10 + delta1 with no sign change
   B. Delta10 + f11 X1 = delta1
   C. Delta10 X1 + f11 = delta1
   D. Delta10 - f11 = X1
   Answer: B. By superposition the displacement at the release is Delta10 + f11 X1, and it must equal the actual displacement delta1. (Key point: Derivation)

3. For the hinged portal with unit horizontal force at D, f11 = (1/EI)[2(6^3/3) + 6^2(3)]. What does each term represent?
   A. Beam term 144 and column term 108
   B. Both terms are columns
   C. Columns 144 (m = y) and beam 108 (m = 6 constant)
   D. Columns 108 and beam 144
   Answer: C. Each column has m = y giving 6^3/3 = 72 (two: 144); the beam has m = 6 over 3 m giving 108. (Key point: Flexibility coefficient)

4. In the beam, the mean temperature 15°C and the gradient 10°C give which effects on the displacement of D?
   A. Expansion moves D outward by 0.495 mm; curvature moves it inward by 3.3 mm, net inward
   B. Only expansion matters
   C. Curvature moves D outward
   D. Both effects move D outward
   Answer: A. Axial term alpha*tm*L = 4.95e-4 m outward; curvature term -3.3e-3 m inward; net -2.805e-3 m. (Key point: Thermal displacement)

## Related in Theory of Structures II

- [Compatibility equation for two-hinged arch](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-compatibility-equation) — Theory of Structures II, chapter 2
- [Flexibility matrix method: portal frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aflexibility-matrix-portal-12000) — Theory of Structures II, chapter 2
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- [Force method: frame with hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-30knm-50kn-hinged) — Theory of Structures II, chapter 2
- [Force method: fixed portal frame, 30 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aportal-fixed-force-method-4x4-30knm) — Theory of Structures II, chapter 2
- [Force method: frame reactions at hinged support](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-50knm-hinged-d) — Theory of Structures II, chapter 2

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