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3 years ago

A driver on a leveltwo-lane highway observes a

Engineering

1 answer:

NARA [144]3 years ago

3 0

Answer:

SSD = 565.54 ft

Explanation:

Given data:

driver speed = 60 mph

clear distance available to stop the truck on emergency can be calculated by using stopping sight distance

SSD = Lag distance + breaking distance

$SSD = vt + \frac{v^2}{2gf}$

let take comfortable acceleration rate as a = 11.2 ft/s^2

g =32.2 ft/s^2

a = fg

$f = \frac{11.2}{32.2} = 0.348$

speed is given as 60 mph

we know that 1 mile 5290 ft

i hr = 3600 sec

therefore v = 88 ft./s

from standard assumption time to reaction is taken as 2.5 sec

so $SSD = 88 \times 2.5 + \frac{88^2}{2\times 32.3\times 0.348}$

SSD = 565.54 ft

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3 years ago

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A Simply supported wood beam with overhang is subjected to uniformly distributed load q. The beam has a rectangular cross sectio

irinina [24]

Answer:

q = 61.71 KN/m

Explanation:

We know that shear force at one end of the beam is;

F = wl/2

Where;

w is the uniformly distributed load and l is the span.

Thus, in this question, q is the distributed load, so;

F = ql/2

Area of beam section = breadth x depth

In this case,

Area = 200 × 250 = 50000 mm²

We are given allowable shear stress of τa=1.8MPa. This can also be written as τa = 1.8 N/mm²

We know that formula for average shear stress is;

τ_avg = Force/Area

Thus, Force = τ_avg x Area

However, we are given maximum allowable shear stress as 1.8and we know that; τ_max = 1.5 × τ_avg

Thus, τ_avg = 1.8/1.5 = 1.2

Hence;

Force = 1.2 × 50,000 = 60000 N

We need

So from the earlier equation F = ql/2,we can get; 60000 = ql/2

ql = 120000 - - - - - (1)

Now, to the bending stress, we know that section modulus of a rectangular section is;

Z = bd²/6

So,for this question, we have;

Z = (200 × 250²)/6

Z = 2083333.33 mm²

Maximum bending moment of a simply supported beam is wl²/8

So,in this case, M = ql²/8

So,formula for maximum bending stress = M/Z

So, plugging in the values, we have ;

σ_max = (ql²/8) / 2083333.33

We are given σ= 14 MPa or 14 N/mm²

Thus;

14 = (ql²/8) / 2083333.33

ql² = 14 × 2083333.33 × 8

ql² = 233333332.96 - - - eq(2)

From equation 1,we saw that;ql = 120000.

Putting this for ql in equation 2,we will get;

120000l = 233333332.96

l = 233333332.96/120000

l = 1944.44 mm

So from eq 1,q = 120000/l

q = 120000/1944.44

q = 61.71 KN/m

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3 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

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Answer:

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Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

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solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

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$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

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