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# Elasticity Elements – Stress Functions

The stress function was introduced by __________

a) G.B Airy

b) Terzaghi

c) Darcy

d) Meyerhof

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Answer: a

Explanation: The stress function was introduced by G.B Airy in 1862. The stress function is denoted by Φ and is called Airy stress function. Darcy gave the law of flow of water through soils.

The σ_{x} in terms of stress function is given by __________

a) ∂Φ∂z

b) ∂2Φ∂z2

c) ∂2Φ∂x2

d) ∂Φ∂x

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Answer: b

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.

The stress function for stress in the x-direction is given by,

∂2Φ∂z2.

The σ_{z} in terms of stress function is given by __________

a) ∂Φ∂z

b) ∂2Φ∂z2

c) ∂2Φ∂x2

d) ∂Φ∂x

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Answer: c

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.

The stress function for stress in the z-direction is given by,

σz=∂2Φ∂x2.

The shear stress τ_{xz} in terms of stress function is given by __________

a) ∂Φ∂z−γx

b) −∂2Φ∂x∂z−γx

c) ∂2Φ∂x2−γx

d) ∂Φ∂x−γx

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Answer: b

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.

The shear stress function τ_{xz} is given by,

τxz=−∂2Φ∂x∂z−γx

For both plane stress as well as plain strain case the equilibrium equation in x-direction is _______

a) ∂σx∂x+∂τyx∂y=0

b) ∂σx∂x+∂τyx∂y+∂τzx∂z+X=1

c) ∂σx∂x+∂τyx∂y+X=0

d) ∂σx∂x+∂τzx∂z=0

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Answer: d

Explanation: The equilibrium equation in x-direction is ∂σx∂x+∂τyx∂y+∂τzx∂z+X=0

In the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.

∴ The equation reduces to ∂σx∂x+∂τzx∂z=0

For both plane stress as well as plain strain case the equilibrium equation in z-direction is _______

a) ∂τxz∂x+∂σz∂z+γ=0

b) ∂σx∂x+∂τzx∂z+γ=1

c) ∂σx∂x+∂τyx∂y+γ=0

d) ∂σx∂x+∂τzx∂z=0

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Answer: a

Explanation: The equilibrium equation in x-direction is ∂τxz∂x+∂τyz∂y+∂σz∂z+γ=0.

In the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.

∴ The equation reduces to ∂τxz∂x+∂σz∂z+γ=0

For two dimensional case, for both plane stress as well as plain strain case the compatibility equation is _______

a) ∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z

b) ∂2εz∂z2+∂2εy∂x2=∂2Γzy∂z∂y

c) ∂2εx∂y2+∂2εy∂x2=∂2Γxy∂x∂y

d) ∂2εz∂z2+∂2εy∂x2=0

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Answer: a

Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;

∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z.

This is because, in the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.

The compatibility equation in terms of plane stress case is given by ________

a) (∂2∂x2+∂2∂z2)=0

b) (∂2∂y2+∂2∂z2)(σy+σz)=0

c) (∂2∂x2+∂2∂y2)(σx+σy)=0

d) (∂2∂x2+∂2∂z2)(σx+σz)=

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Answer: d

Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;

∂2εx∂z2+∂2εz∂x2=∂2Γxz∂x∂z———————(1)

From the Hooke’s law equation,

εx=1E(σx−μσz),

εz=1E(σz−μσx) and

Γxz=2(1+μ)Eτzx

Substituting these values in (1) and simplifying further we get,

(∂2∂x2+∂2∂z2)(σx+σz)=0.