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# Stress Distribution – Triangular Loadings

The uniformly varying load is __________ in a beam.

a) rate of loading increases linearly from zero

b) rate of loading increases non-linearly from zero

c) equal load at every point

d) equal load only at supports

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Answer: a

Explanation: The uniformly varying load is the rate of loading which increases linearly from zero at one end of the support.

The triangular load is also known as ___________

a) uniformly distributed load

b) uniformly varying load

c) point load

d) equivalent uniformly distributed load

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Answer: b

Explanation: The triangular load is also known as uniformly varying load. The uniformly varying load is the rate of loading which increases linearly from zero at one end of the support.

For any position of point P subtending angle α with AB, the vertical stress is given by___________

a) σz=qaπ[xα(x−α)]

b) σz=qaπ

c) σz=qaπ[xα−az(x−α)2+z2(x−α)]

d) σz=[xα−az(x−α)2+z2(x−α)]

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Answer: c

Explanation: For any position of point P subtending angle α with AB, the vertical stress is given by,

σz=qaπ[xα−az(x−α)2+z2(x−α)]

Where, σ_{z}= vertical stress

x=distance from the support A

z=depth from the ground surface

q=load applied.

For point P under the support A, the vertical stress is given by __________

a) σz=qaπ[xα(x−α)]

b) σz=qaπ

c) σz=qπ[aza2+z2]

d) σz=[xα−az(x−α)2+z2(x−α)]

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Answer: c

Explanation: For point P under the support A, the vertical stress is given by,

σz=qπsin2αA2

∴ σz=qπ[aza2+z2].

For point P under the support B, the vertical stress is given by __________

a) σz=qaπ[xα(x−α)]

b) σz=qπαB

c) σz=qπ[aza2+z2]

d) σz=[xαB−az(x−αB)2+z2(x−αB)]

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Answer: b

Explanation: For point P under the support B, the vertical stress is given by,

σz=qπαB

Where α_B=angle subtended from point P below the support B to the line AB

q=load applied.

For a linearly variable infinite load, for a point P, the vertical stress σ_{z} is _________

a) σz=qaπ[xα+z]

b) σz=qaπ

c) σz=qπ[aza2+z2]

d) σz=[xα−az(x−α)2+z2(x−α)]

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Answer: a

Explanation: let the load intensity be q at a horizontal distance a and the load intensity at distance x is qx=q*x/a.

∴ for any point P the load is given by, σz=qaπ[xα+z].

For a symmetrically distributed triangular load, under the centre of the triangular load, the vertical stress at any point at a depth z is given by ___________

a) σz=qaπ[α1+α2]

b) σz=qπ[α1+α2]

c) σz=qπ[aza2+z2]

d) σz=qπ[α1−α2]

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Answer: b

Explanation: Considering two triangular loads of AOC and OCB separately and adding them we get,

σz1=qπα1 for triangular load of AOC

σz2=qπα2 for triangular load of OCB

∴ σz=σz1+σz2

∴ σz=qπ[α1+α2].

For a symmetrically distributed triangular load, the shear stress τ_{xz} at any point at a depth z is given by ___________

a) τxz=−qzaπ[α1−α2]

b) τxz=−qπ[α1+α2]

c) τxz=−qπ[aza2+z2]

d) τxz=−qπ[α1−α2]

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Answer: a

Explanation: For a symmetrically distributed triangular load, the shear stress τ_{xz} at any point at a depth z is given by,

τxz=−qzaπ[α1−α2]where α_{1} and α_{2} are angles subtended by the point P at AO and BO respectively

z=depth.

For a triangular and uniformly distributed semi-infinite loads, the shear stress τ_{xz} in the plane xz is ___________

a) τxz=−qzaπα

b) τxz=−qπα

c) τxz=−qπ[aza2+z2]

d) τxz=−qπz

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Answer: a

Explanation: For a triangular and uniformly distributed semi-infinite loads, the shear stress τ_{xz} in the plane xz is given by,

τxz=−qzaπ α where α is the angle subtended by point P on OA,

z=depth

q=load intensity.

For a load intensity of q=20kN/m, find the shear stress τ_{xz} at a depth 5m from the given diagram.

a) -5.2 kN/m^{2}

b) -6.2 kN/m^{2}

c) -7.2 kN/m^{2}

d) -8.2 kN/m^{2}

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Answer: b

Explanation: Given,

Load q=20kN/m

α=30°=0.523rad

a=2.68m

z=5m

∴ τxz=−qzaπα

τxz=−20∗52.68∗π0.523

τ_{xz}=-6.2 kN/m^{2}.

For a triangular and uniformly distributed semi-infinite loads, the vertical stress σ_{z} is given by ___________

a) σz=qaπ[xα+z]

b) σz=qaπ(aβ+xα)

c) σz=qπ[aza2+z2]

d) σz=[xα−az(x−α)2+z2(x−α)]

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Answer: b

Explanation: For a triangular and uniformly distributed semi-infinite loads, the vertical stress σ_{z} is given by,

σz=qaπ(aβ+xα) where α is the angle subtended by point P on OA,

β= angle at point P between horizontal line and PA

q=load intensity.

For a load intensity of q=20kN/m, find the vertical stress σ_{z} from the given diagram.

a) 5.62 kN/m^{2}

b) 6.23 kN/m^{2}

c) 13.33 kN/m^{2}

d) 8.32 kN/m^{2}

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Answer: c

Explanation: Given,

Load q=20kN/m

α=60°=1.047rad

β=60°=1.047rad

a=2.68m

the vertical stress σ_{z} is,

σz=qaπ(aβ+xα)

σz=202.68∗π(2.68∗1.047+2.68∗1.047)

∴ σ_{z}=13.33 kN/m^{2}.

Trapezoidal load is encountered in earth fills.

a) True

b) False

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Answer: a

Explanation: Trapezoidal loadings are quite commonly encountered in earth fills, embankments, highways, rail road, earth dams, etc. The trapezoidal loadings can be split into a triangular loading minus another triangular loading of smaller magnitude.

The maximum shear stress is the difference between major and minor principal stresses.

a) True

b) False

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Answer: b

Explanation: The maximum shear stress is half the difference between major and minor principal stresses.

τmax=12(σ1−σ2).

The greatest value of maximum shear stress τ_{max} occurs when angle θ is _________

a) π

b) π/2

c) π/3

d) π/4

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Answer: b

Explanation: The maximum shear stress is given by,

τmax=12(σ1−σ2)=qπsinθ

Sinθ is maximum when θ= π/2

∴ Sin π/2=1

∴ τmax=qπ.