Engineers need to look for non-technological systems, as well as technological problems, to solve problems. true or false

A material point in equilibrium has 1 independent component of shear stress in the xz plane. a)True b)- False

ozzi

Answer:

True

Explanation:

For point in xz plane the stress tensor is given by $\left[\begin{array}{ccc}Dx_{} &txz\\tzx&Dz\\\end{array}\right]$

where Dx is the direct stress along x ; Dz is direct stress along z ; tzx and txz are the shear stress components

We know that the stress tensor matrix is symmetrical which means that tzx = txz ( obtained by moment equlibrium )

thus we require only 1 independent component of shear stress to define the whole stress tensor at a point in 2D plane

8 0

3 years ago

In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had

kipiarov [429]

Answer:

Moisture/ water content w = 26%

Void ratio , e = 0.73

Explanation:

Initial mass of saturated soil w1 = mass of soil - weight of container

= 113.27 g - 49.31 g = 63.96 g

Final mass of soil after oven w2 = mass of soil - weight of container

= 100.06 g - 49.31 g = 50.75

Moisture /water content, w = $\frac{w1-w2}{w2}$ = $\frac{63.96-50.75}{50.75}$ = 0.26 = 26%

Void ratio = water content X specific gravity of solid

= 0.26 X 2.80 =0.728

5 0

3 years ago

Select the correct answer. Which existing technology did engineers use to enhance the speed of propeller-driven airplanes

Musya8 [376]

metallurgy:

Explanation:

7 0

3 years ago

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete

Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making Y the subject

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}$

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, $\sigma_c$ is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

$Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682$

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

$\sigma_c=\frac {K}{Y \sqrt {a\pi}}$ and making K the subject

$K=\sigma_c Y \sqrt {a\pi}$ and substituting 260 MPa for $\sigma_c$ while a is taken as 0.003m and Y is already known

$K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa$

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0

3 years ago

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at

makkiz [27]

Answer:

$\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}$