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2 years ago

. A normal-weight concrete has an average compressive strength of 20 MPa. What is the estimated flexure strength

Engineering

1 answer:

bulgar [2K]2 years ago

4 0

Answer:

2.77mpa

Explanation:

compressive strength = 20 MPa. We are to find the estimated flexure strength

We calculate the estimated flexural strength R as

R = 0.62√fc

Where fc is the compressive strength and it is in Mpa

When we substitute 20 for gc

Flexure strength is

0.62x√20

= 0.62x4.472

= 2.77Mpa

The estimated flexure strength is therefore 2.77Mpa

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3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

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a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

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$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

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$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$

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