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liq [111]
3 years ago
6

Why is concrete on its own not a good material to use

Engineering
1 answer:
Inga [223]3 years ago
7 0

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

Concrete exhibits high compressive strength when used. However, it has very low compressive strength. This is the reason why concrete is normally combined with steel to make a composite building material called reinforced concrete. The steel reinforces concrete hence increasing the tensile strength in RC buildings. The end composite is durable and fireproof. Generally, the main reason why concrete is not use on its own is due to its poor tensile strength.

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Need help fast I been stuck in this for the longest
gavmur [86]

Answer:

3rd and 4rth

Explanation:

a geologist studies the earth and both of these have something to do with the earth

6 0
3 years ago
A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure
aleksandr82 [10.1K]

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of ​​the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N                       [n = Bronze]

Pₓ =  3924 N                             [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

#SPJ9

4 0
1 year ago
The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th
Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}

Given that t= 200 days

R(200)=e^{-\left(\dfrac{200}{300}\right)^2}

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

0.95=e^{-\left(\dfrac{t}{300}\right)^2}

by solving above equation t=67.74 days

When R=90 %

0.90=e^{-\left(\dfrac{t}{300}\right)^2}

by solving above equation t=97.2 days

7 0
3 years ago
Is Micah a idiot true or false ooooooooooooooooooooooooooooooooooo
Vedmedyk [2.9K]
False because yeah jkdkdlgkdjfkekvkx
6 0
3 years ago
Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)
ArbitrLikvidat [17]

Answer:

\mathbf{P_x =25 \ watts}

\mathbf{x_{rmx} = 5 \ unit}

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) +  sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal  Asin (ωt), Power can be expressed as:

P= \dfrac{A^2}{2}

For the number of sinosoidial signals;

Power can be expressed as:

P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...

As such,

For x(t), Power  P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}

P_x = \dfrac{25}{2}+ \dfrac{25}{2}

P_x = \dfrac{50}{2}

\mathbf{P_x =25 \ watts}

For the number of sinosoidial signals;

RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...

For x(t), the RMS value is as follows:

x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }

x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }

x_{rmx }=\sqrt{(\dfrac{50}{2} )}

x_{rmx} =\sqrt{25}

\mathbf{x_{rmx} = 5 \ unit}

8 0
3 years ago
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