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Rina8888 [55]
3 years ago
13

Consider the following example: The 28-day compressive strength should be 4,000 psi. The slump should be between 3 and 4 in. and

the maximum aggregate size should not exceed 1 in. The coarse and fine aggregates in the storage bins are wet. The properties of the materials are as follows:________.
Cement : Type I, specific gravity = 3.15
Coarse Aggregate: Bulk specific gravity (SSD) = 2.70; absorption
capacity = 1.1%; dry-rodded unit weight = 105 lb./ft.3
surface moisture = 1%
Fine Aggregate: Bulk specific gravity (SSD) = 2.67; absorption
capacity = 1.3%; fineness modulus = 2.70;
surface moisture = 1.5%
Engineering
1 answer:
lara [203]3 years ago
8 0
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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
3 years ago
A steam power plant is represented as a heat engine operating between two thermal reservoirs at 800 K and 300 K. The temperature
Sergeeva-Olga [200]
Yeet is the answer .....
4 0
2 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
The purpose of pasteurizing milk is to
katen-ka-za [31]

Answer:

i think it c

Explanation:

6 0
3 years ago
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Thermosets burn upon heating. a)-True b)- false?
hram777 [196]

Answer:

true

Explanation:

True, there are several types of polymers, thermoplastics, thermosets and elastomers.

Thermosets are characterized by having a reticulated structure, so they have low elasticity and cannot be stretched when heated.

Because of the above, thermosetting polymers burn when heated.

5 0
3 years ago
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