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3 years ago

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A rectangular beam having b=300 mm and d=575 mm, spans 5.5 m face to face of simple supports. It is reinforced for flexure with

4φ32 bars that continue uninterrupted to the ends of the span. It is to carry a service dead load wD=30 kN/m (including self-weight) and a service live load =45 kN/m both uniformly distributed along the span. Design the shear reinforcement using φ10 vertical U stirrups. Use the equation (a) for Vc. Material strengths are fc’=22 and fy=420 MPa

1 answer:

katovenus [111]3 years ago

8 0

Answer:

provide 180 mm spacing

Explanation:

GIVEN DATA

rectangular beam: (b) = 300 mm, (d) = 575 mm

reinforced for flexure = 4Ф32 bars

WD = 30 kN /m, WL = 45 kN/m

Wu = 1.4 * 30 + 1.6 * 45 = 114 kN/m

i) concrete shear stress ( vc)

100 Ac / bd = (100 * u * $\frac{\pi }{4}$ * 32^2) / 300 * 575 = 1.865

from table 3.8

when: 100 Ac / bd = 1.865 then Vc = 0.778 N/mm^2

Ultimate shear force = (114 *5.5) / 2 = 313.5 kN

design shear stress = V / bd = (313.5 * 10^3) / (300 * 575) = 1.82 N/mm^2

v < 0.8 $\sqrt{22}$ = 1.82 < 3.75

design link provided according to

Asv / sv = b(v-vc) / 0.87 fy = 300(1.82 - 0.778) / 0.87 (420)

ASv / Sv = 0.855

From table 3.13 :the value of Asv / sv can be calculated as

$\frac{0.855 - 0.785}{0.897 - 0.785} = \frac{x - 200}{175 - 200}$

x = (-25) [ 0.625] + 200 = 184.375 mm

provide 180 mm spacing

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Answer:

The mass composition of ∝ ferrite is 88.94%.

The mass composition of cementite is 11.05%.

Explanation:

Given that

T=726 °C

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Lets take data from ideal Iron -carbon diagram at 726 °C

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We know that if we want to find the mass fraction the we use Lever rule .So now by using lever rule

$The\ mass\ composition\ of\alpha \ ferrite =\dfrac{6.7-0.76}{6.7-0.022}$

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2 years ago

The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h

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Answer:

$M=0.0411 kg/h or 4.1*10^{-2} kg/h$

Explanation:

We have to combine the following formula to find the mass yield:

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Now substitute the values

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2 years ago

Vehicles arrive at a single toll booth beginning at 8:00 A.M. They arrive and depart according to a uniform deterministic distri

jok3333 [9.3K]

Answer:

Explanation:

answers attached bellow

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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.

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A) DF=0.17

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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:

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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:

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

This material will break.

3 0

2 years ago

Two kilograms of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m3 and 1 bar, re

valentinak56 [21]

Answer: Heat transfer (Q) is 521 kJ.

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V is volume;

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As pressure is constant:

$\frac{V_{i} }{T_{i} } =\frac{V}{T}$

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Q = 521 kJ

The heat transfer in the process is 521 kJ.

6 0

2 years ago

Read 2 more answers