How is a scale model different from other types of models?

Ring rolling is a deformation process in which a thick-walled ring of smaller diameter is rolled into a thin-walled ring of larg

VMariaS [17]

Answer:

a)True

Explanation:

Rolling:

Rolling is a metal forming in which a material passes through two or more than two depends on conditions,rolls to produce the desired product.

Ring rolling:

In ring rolling a thick ring compresses by rolls to produce the large diameter ring.Actually volume of material is constant so when diameter of ring increases then to compensate it, the thickness of ring reduces .In simple words we can say that in ring rolling a thick ring of smaller is rolled into a thin ring of larger diameter.

6 0

3 years ago

Could I please get help with this

alex41 [277]

Answer:

1. $I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. $I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = 37. $\overline 3$ in.⁴

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = 149. $\overline 3$ in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458 $\overline 3$

$I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458 $\overline 3$

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458 $\overline 3$

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548 $\overline 3$

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 4 × 7³/12 = 114. $\overline 3$

$I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 7 × 4³/12 = 37. $\overline 3$

$I_{yc}$ = 37. $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457. $\overline 3$

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 149. $\overline 3$

$I_y$ = 149. $\overline 3$ in.⁴

3. The deflection, $\delta _{max}$ , of a simply supported beam having a point load at the center is given as follows;

$\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}$

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, $I_{xc}$ = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

$\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552$

The maximum deflection of the beam, $\delta _{max}$ = 2.55552 inches

5 0

3 years ago

In which of the following states are you most likely to find a home with a basement and why?

seraphim [82]

Well a home with a basement maybe a haunted house or medium house and bc that houses like that are still known today but I would say a medium house.

6 0

3 years ago

If x < 5 and x >c, give a value of c such that there

Arlecino [84]

we have

x<5

x>c

we know that

The solution is the intersection of both solution sets of the given inequalities.

The solutions of the compound inequality must be solutions of both inequalities.

The value of c could be 5 or any number greater than 5, such that there are no solutions to the compound inequality

Because

A number cannot be both less than 5 and greater than 5 at the same time

therefore

the answer is

for c_> there are no solutions to the compound inequality

7 0

3 years ago

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$$P(h$

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore, $x_1=368 \ veh/h$

$=\frac{368}{1000} = 0.368$

Given, $t_1=2+x_1$

= 2 + 0.368

= 2.368 min

At user equilibrium, $t_2=t_1$

∴ $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

= 1368 veh/h

7 0

3 years ago