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3 years ago

Why do we care about a material's ability to resist torsional deformation?

Engineering

1 answer:

lesya692 [45]3 years ago

5 0

Answer:

(A) Because the angle of twist of a material is often used to predict its shear toughness

Explanation:

In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.

The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.

The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:

1- Tangential tensions appear parallel to the cross section.

2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.

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Singularity is an important property of a square matrix. This is also known as degenerate. What is the value of the determinant

DENIUS [597]

Answer:

For a Singular matrix, the determinant must be equivalent to 0.

Explanation:

A matrix is a rectangular array in which elements are arranged in rows and columns.

Each square matrix has a determinant. The determinant is a numerical idea that has a fundamental function in finding the arrangement just as investigation of direct conditions. For a Singular matrix, the determinant must be equivalent to 0.

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3 years ago

Impedance is defined as the total opposition to current in an AC circuit. Question 17 options: True False

ArbitrLikvidat [17]

Answer: True

Explanation:

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2 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$