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3 years ago

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A girder that carries a uniformly distributed dead load of 1.7 k/ft plus its self-weight and three concentrated live loads of 15

kips at the quarter points of the 36 ft span is to be sized. Using A992 steel, determine the lightest Wshape to carry the load with lateral supports provided at the supports and load points. Use the correct Cb, check shear, and limit deflection to 1/360 of span.

1 answer:

posledela3 years ago

3 0

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A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w

Verdich [7]

Explanation:

Outer di ameter $d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}$

$\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10$

$d_{1}=380 \mathrm{mm}$

Given loading on the cylinder $P=300 \mathrm{kN}$ Helix an gle of the weld form $\theta=20^{\circ}$

(i) Normal stress on the plane at angle $\theta=20^{\circ}$ is

$\sigma=\frac{P \cos ^{2} \theta}{A_{0}}$

$\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)$

$\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)$

$=12252.21 \mathrm{mm}^{2}$

$=12.25221 \times 10^{-9} \mathrm{m}^{2}$

$\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}$

$=-21.6 \mathrm{MPa}$

(ii) Shear stress along an angle of $\theta=20^{\circ}$ is $\tau=\frac{P}{A_{0}} \cos \theta \sin \theta$

$=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}$

$=-7.86 \mathrm{MPa}$

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3 years ago

Why is the face of the claw on a claw hammer usually a smooth curve? Why isn't it straight or some other shape?

GarryVolchara [31]

Answer:

The face of the claw on the claw hammer is usually a smooth curve so as to improve the ease with which nails are removed when removing nails because as the nail held between the V shaped split claw is being pulled out from the wood, it slides more and more towards cheek, reducing the distance of the nail from the cheek which is the fulcrum, thereby increasing the mechanical advantage because the location of the hand on the grip remains unchanged

Explanation:

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3 years ago

Any programmer who writes a Diophantine equation solver must occasionally encounter an infinite loop.

Sergio [31]

Answer: True

Explanation:

An Infinite loop occurs in a computer programming when a sequence of instructions run endlessly without stopping until there is an external intervention, this intervention could be pull or plug. It is also known as endless loop. It occurs due to using of variables that are not properly updated or when there is an error in looping condition.

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3 years ago

IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The follow

Irina-Kira [14]

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, $h_0=2.2$ inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

$\Delta h = H_0-h_f=\mu^2R$

$=0.2^2 \times 14 = 0.56$ inches

$h_f = 2.2 - 0.56$

= 1.64 inches

Roll strip contact length (L) = $\sqrt{R(h_0-h_f)}$

$=\sqrt{14 \times 0.56}$

= 2.8 inches

Absolute value of true strain, $\epsilon_T$

$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$

Average true stress, $\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$ Psi

Roll force, $L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$

= 788,900 lb

For SI units,

Power = $\frac{2 \pi FLN}{60}$

$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$

= 10399.81168 W

Horse power = 13.9357

6 0

2 years ago

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture tough

sp2606 [1]

Answer:

Not subject to detection

Explanation:

Assuming the value of strain fracture toughness is 77 Mpa $\sqrt m$

The design stress is half hence $\sigma=0.5\times 1400=700 Mpa$

Critical flaw size, $a_c=\frac {1}{\pi}(\frac {K_{1c}}{Y/sigma})^{2}$

Where Y is dimensionless parameter, $\sigma$ is applied stress, $K_{1c}$ is plane strain fracture toughness, $a_c$ is critical length of surface crack

$a_c=\frac {1}{\pi}(\frac {77}{1*700})^{2}= 0.0038515496\approx 0.00385m$

The critical length of surface crack is therefore 3.85 mm, which is less than detection apparatus size given as 4 mm

Since the critical flaw size is less than the resolution limit of flaw detection apparatus, the critical flaw for this plate is not subjected to detection.

5 0

2 years ago