Can someone help me plz!!

Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l

ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0

3 years ago

Read 2 more answers

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Genrish500 [490]

Answer:

ummm why is you doing this

Explanation:

It doesnt make sense.

7 0

3 years ago

Drag the tiles to the boxes to form correct pairs. Identify the designations of the three employees in an automobile company fro

aniked [119]

Answer:

is the fare of our responsibility towards

7 0

3 years ago

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal

7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )

m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04

now lets find the value of h/mk

h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb = (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]

so we substitute

0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0

3 years ago

An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i

Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object $\left ( m\right )$ =521kg

initial velocity $\left ( v_0\right )$ =90m/s

Final velocity $\left ( v\right )$ =14m/s

kinetic energy of body is given by= $\frac{1}{2}$ $m$ $v^{2}$

change in kinectic energy is given by substracting final kinetic energy from initial kinetic energy of body.

Change in kinetic energy= $\frac{1}{2}\times m$ $\left ( V_0^{2}-V^2\right )$

Change in kinetic energy= $\frac{1}{2}\times521$ $\left ( 90^{2}-14^2\right )$

Change in kinetic energy=2058.992KJ

7 0

3 years ago