Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337385497
Author: WRIGHT, Wendelin J.
Publisher: Cengage,
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Chapter 6, Problem 6.29P
Interpretation Introduction
Interpretation:
The dimensions of the polymer bar are 1in× 2in×15in. The modulus of elasticity of bar is 600,000psi. Value of force which is required to stretch the bar from 15 into 15.25 in is to be determined.
Concept introduction:
The concept of strain, stress and Hooke's law will be used. The formula of strain and Hooke's law is given below -
The relation between true strain and engineering strain is given as-
And Hooke's law-
Expert Solution & Answer
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Write handwritten solution, answer a,b and c
Refer to the soil profile shown in the Figure
a. Calculate the variation of o, u, and o' with depth.
b. If the water table rises to the top of the ground surface, what is the change in the effective stress at the bottom of the clay layer?
c. How many meters must the groundwater table rise to decrease the effective stress by 15 kN/m? at the bottom of the clay layer?
Water is discharged into the atmosphere
through a bent nozzle of an angle (a) as
shown in the figure. The cross-sectional
area at the nozzle inlet and outlet are (Ain)
and (Aout), respectively. The discharge
through the nozzle is (Q). The gauge
pressure at the nozzle inlet is (Pin). The
bend lies in a horizontal plane.
Vin
Ain
Aout
Atmosphere
Vout
Problem (9): Given the values of Ain [m²], Aout [m²], Pin [atm], Q [m³/s], and a [degrees], calculate the
magnitude of the reaction force component in x-direction (Rx) in [N].
Givens:
A in = 0.301 m^2
Aout
Pin
=
0.177 m^2
1.338 atm
Q
α
=
0.669 m^3/s
37.183 degrees
Answers:
( 1 ) 23028.076 N
( 2 ) 29697.962 N
( 3 ) 18633.611 N
( 4 ) 14114.988 N
Please answer the following question in the picture and show all of your work please.
Chapter 6 Solutions
Essentials Of Materials Science And Engineering
Ch. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10P
Ch. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.20PCh. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.23PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - Prob. 6.35PCh. 6 - Prob. 6.36PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38PCh. 6 - Prob. 6.39PCh. 6 - Prob. 6.40PCh. 6 - Prob. 6.41PCh. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - Prob. 6.52PCh. 6 - Prob. 6.53PCh. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Prob. 6.57PCh. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - Prob. 6.60PCh. 6 - Prob. 6.61PCh. 6 - Prob. 6.62PCh. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Prob. 6.72PCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. K6.1KP
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- Please answer the following questions and make sure you answer each question please.arrow_forwardWater is discharged into the atmosphere through a bent nozzle of an angle (a) as shown in the figure. The cross-sectional area at the nozzle inlet and outlet are (Ain) and (Aout), respectively. The discharge through the nozzle is (Q). The gauge pressure at the nozzle inlet is (Pin). The bend lies in a horizontal plane. Ain Vin Aout X Atmosphere Vout Problem (10): Given the values of Ain [m2], Aout [m²], Pin [atm], Q [m³/s], and a [degrees], calculate the magnitude of the reaction force component in y-direction (Ry) in [N]. Givens: A in 0.169 m^2 A out Pin 0.143 m^2 0.552 atm = Q α 0.367 m^3/s = 31.72 degrees Answers: ( 1 ) 6264.193 N (2) 12041.886 N ( 3 ) 8715.747 N ( 4 ) 7139.937 Narrow_forwardProblem (12): A pump is being used to lift water from the bottom tank to the top tank in a pipe of diameter (d) at a discharge (Q). The pipe system comprises four Long radius 90° threaded elbows. The pipe entrance is sharp-edged, and the pipe exit is sudden. A Ball valve (1/3 closed) is used to control the discharge in the pipeline. Given the values of Q [Lit/s], and d [cm], calculate the power loss due to components (i.e., minor losses) in the pipe (Wminor-loss) in [W]. Givens: Q = 12.275 lit/s d = 6.266 cm Answers: ( 1 ) 1142.006 W (2) 952.086 W ( 3 ) 1225.555 W ( 4 ) 1331.216 W Loss Coefficients for Pipe Components (h,= K,Y) Component a. Elbows KL elbow Regular 90°, flanged 0.3 Regular 90°, threaded 1.5 Long radius 90°, flanged 0.2 V 90° elbow Long radius 90°, threaded 0.7 Long radius 45°, flanged 0.2 0.4 Regular 45°, threaded . 180° return bends 180° return bend, flanged 0.2 V 45° elbow 180° return bend, threaded 1.5 c. Tees Line flow, flanged 0.2 Line flow, threaded 0.9 180°…arrow_forward
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