(True/False) Unix is written in the C language. * True O False

Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display.

creativ13 [48]

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740 × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

8 0

3 years ago

Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d

navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

7 0

3 years ago

Read 2 more answers

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

$0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}$

E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0

2 years ago

1. What affects the weight of an airplane? a. Mass b. Matter c. Materials d. Speed

BARSIC [14]

Mass affects the weight of an airplane because everything has mass

5 0

2 years ago

Read 2 more answers

How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work

lozanna [386]

Answer:

its so simple. u must mind some formulas.

Explanation:

mili->10^(-3)

micro->10^(-6)

nano->10^(-9)

so write the exact number and move "." to left or right depend question.

in this one:

23.5 is 23500000.0 nano with a default dot at the end

for turning to mili u must move the dot 6 steps to left so it will be: 23.5 milimeter.

6 0

3 years ago