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Civil

Axial stiffness of a diagonal brace against storey sway

Horizontal stiffness that one pin-ended diagonal brace of length L, modulus E and area A gives to a floor, with θ the brace angle to the horizontal.

Ref: CIV-37
Rev: v1
Field: Civil
Result: K_lat [N/m]

K_{lat} = \frac{E A^{\cos 2} θ}{L}

A diagonal brace at angle θ to the horizontal that is pulled by a small horizontal floor movement δ stretches by δ cos θ, so its axial force is (EA/L)δ cos θ. The horizontal component of that force is another factor cos θ, so the lateral stiffness is EA cos²θ/L. A brace at 45° has half the axial stiffness as lateral stiffness, and one at 90° has none.

Horizontal stiffness that one pin-ended diagonal brace of length L, modulus E and area A gives to a floor, with θ the brace angle to the horizontal. A small horizontal translation stretches the brace by cos θ per unit, and the horizontal component of its force adds another cos θ. A compression brace must remain stable.

Calculator

Solve for

K_latLateral stiffnessComputed

N/m

EModulus

1e104e11

ABrace area

m²

1e-40.01

LBrace length

120

θAngle to the horizontal

rad

0.11.5

Axial stiffness of the brace, EA/L4×10⁷ N/m
Efficiency, cos²θ0.64
Floor displacement for a 10 kN horizontal force0.3906 mm

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Lateral stiffness (N/m) current value 2.56e7 N/m

ABrace areaKey driver

2,560,0002.56e8

LBrace length

6,400,0001.28e8

EModulus

1,280,0005.12e7

θAngle to the horizontal

200,0003.96e7

The brace gives 25.6 MN/m of lateral stiffness, 64 % of its axial value.
A 10 kN horizontal force would sway the floor 0.3906 mm; with a drift limit of 1/400 of a 3.5 m storey (8.75 mm), that is within the limit.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (N/m)	Local step	Change in result
A Brace area	1e-4 to 0.01 m²	2,560,000	2.56e8	2.534e8	+10%	+10 %
L Brace length	1 to 20 m	6,400,000	1.28e8	1.216e8	+10%	−9.09 %
E Modulus	1e10 to 4e11 Pa	1,280,000	5.12e7	4.992e7	+10%	+10 %
θ Angle to the horizontal	0.1 to 1.5 rad	200,200	3.96e7	3.94e7	+10%	−9.81 %

Sweep of Angle to the horizontal

Angle to the horizontal (rad)	Lateral stiffness (N/m)
0.1	3.96e7
0.275	3.705e7
0.45	3.243e7
0.625	2.631e7
0.8	1.942e7
0.975	1.26e7
1.15	6,674,000
1.325	2,368,000
1.5	200,200

Sweep one variable

Vary

Lateral stiffness (N/m)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

K_lat

Lateral stiffness

Result · N/m

What the equation gives you. Everything else feeds it.

Modulus

Pa · typical 1e10 to 4e11

The Young’s modulus of the brace, in Pa. Raises the stiffness in direct proportion. Raising it by 10 % (to 2.2×10¹¹) moves the result by +10 %, from 2.56×10⁷ to 2.816×10⁷ N/m.

Brace area

m² · typical 1e-4 to 0.01

The brace cross-section area, in m². Raises the stiffness in direct proportion. Raising it by 10 % (to 0.0011) moves the result by +10 %, from 2.56×10⁷ to 2.816×10⁷ N/m.

Brace length

m · typical 1 to 20

The brace length, in m. A longer brace is more flexible axially, lowering the stiffness in inverse proportion. Raising it by 10 % (to 5.5) moves the result by −9.09 %, from 2.56×10⁷ to 2.327×10⁷ N/m.

Angle to the horizontal

rad · typical 0.1 to 1.5

The angle of the brace to the horizontal, in radians. A flatter brace is more effective in sway; stiffness falls with cos² as it steepens. Raising it by 10 % (to 0.7079) moves the result by −9.81 %, from 2.56×10⁷ to 2.309×10⁷ N/m.

Worked example with the values above

Axial stiffness: EA/L = 2×10¹¹ × 0.001 / 5 = 4×10⁷ N/m.
Orientation factor: cos²θ = 0.64.
Klat = 2.56×10⁷ N/m.

Assumptions and limits

Pin joints, loads at the joints, small displacements and linear elastic members.
Tension positive in both the real and virtual force systems.

Common mistakes

Using compression positive in one system and tension positive in the other.
Mixing the real force and the virtual force between members.
Forgetting that unused members must have zero real or virtual force.

Statistics and uncertainty

Truss displacements are exact for pin-jointed bars with axial deformation only. Real joints are slightly flexible and members have length tolerances, so measured displacements are usually 5 % to 20 % higher than the pin-jointed prediction.

Where it is used

Deflection and stiffness checks for trusses, braces and frames.
Estimating thermal movement and strain energy of members.

Maths you need

Virtual work: external virtual work equals internal virtual work
Axial deformation: ΔL = NL/EA
Strain energy: the work stored by elastic deformation

Related formulas

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