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anzhelika [568]

2 years ago

15

A beam spans 40 feet and carries a uniformly distributed dead load equal to 2.2 klf (not including beam self-weight) and a live

load equal to 3.6 klf.
The beam is laterally braced only at the supports and at the midpoint of the span.

Select the lightest W-shape that is adequate for flexure.

Clearly state the controlling limit state and be specific.

1 answer:

salantis [7]2 years ago

6 0

Answer:

A beam is a structural member that is subjected primarily to transverse loads ... stress equal to σy, then the section Moment - Curvature (M-φ) response for .... Example 2.2 Design a simply supported beam subjected to uniformly distributed dead ... (live load). • Step I. Calculate the factored design loads (without self-weight).

Explanation:

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Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

A series AC circuit contains a resistor, an inductor of 250 mH, a capacitor of 4.40 µF, and a source with ΔVmax = 240 V operatin

slega [8]

Answer:

Explanation:

Inductance = 250 mH = 250 / 1000 = 0.25 H

capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)

ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A

a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms

b) capacitive reactance = $\frac{1}{2\pi fC}$ = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms

c) impedance = $\frac{Vmax}{Imax}$ = 240 / 0.11 = 2181.82 ohms

7 0

3 years ago

Read 2 more answers

If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with

SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

$\sigma_b = \frac{M}{I} \times y$

putting all value to get bending stress

$\sigma_b = \frac{50.112}{144} \times 3.96$

$\sigma_b = 1.37 psi$

Bending stress at point 3.96 is $\sigma_b$ = 1.37 psi

3 0

3 years ago

Which of the following is typically wom when working in an atmosphere containing dust?

alukav5142 [94]

Answer:

Either D or C

Both of these masks are used for dust, but since half masks are generally cheaper and easier to use, I'd go with C.

If this is correct, I'd appreciate a brainliest.

3 0

2 years ago

Determine the maximum intensity ww of the uniform distributed load that can be applied to the beam without risk of causing the s

AnnyKZ [126]

Answer:

Please see attachment

Explanation:

Please see attachment

3 0

3 years ago