Ask question

Ask question

All categories

3 years ago

11

A three-point bending test is performed on a glass specimen having a rectangular cross section of height 5.3 mm and width 11.6 m

m; the distance between support points is 50 mm.
Required:
a. Calculate the flexural strength (in MPa) if the load at fracture is 283 N. (Round up the final number to integer-no decimal please.)
b. What is the deflection, Δy=__________ mm, for a load of 265 N if the elastic modulus if the glass is 66 GPa?

1 answer:

wolverine [178]3 years ago

4 0

Answer:

a) 65.139 MPa

b) 17490 GPa

Explanation:

a) calculate the flexural strength

we now that for a three-point bending test the bending moment will occur at the midsection

Max stress ( flexural strength ) = MY / I

Max stress can be expressed as ; 3/2 * ( F*l / b*d^2 ) ------- ( 1 )

F ( load at fracture ) = 283

l (distance between support point ) = 50

b = 11.6

d = 5.3

substitute given data into equation 1 above

= 3/2 * (( 283 * 50) /( 11.6 * 5.3^2) )

= 3/2 * ( 14150 / 325.844 )

= 3/2 * 43.425 = 65.139 MPa

B ) Determine deflection ( Δ y )

load = 265 N

elastic modulus = 66 GPa

elastic modulus = tensile stress / tensile strain

Tensile stress = Δ y

strain = load = 66 GPa

therefore Δ y = 265 * 66 = 17490 GPa

You might be interested in

The following data were obtained when a cold-worked metal was annealed. (a) Estimate the recovery, recrystallization, and grain

Oduvanchick [21]

Sorrry needdddd pointssssss

7 0

3 years ago

technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w

expeople1 [14]

Answer:

both

Explanation:

Both the technician are correct, ac generator output can be tested in both ways. The two ways are current output test to check ac generator output. and voltage output test to check output.

7 0

3 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

4 0

3 years ago

(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and pr

Gnoma [55]

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

7 0

3 years ago

Read 2 more answers

Viteza unui mobil care se deplasează cu accelerație constantă crește de la 3,2 m/s la 5,2 m/s în timp de 8 s. Accelerația mobilu

Verizon [17]

Answer:

$a=0.25\ m/s^2$

Explanation:

Initial speed of the mobile = 3.2 m/s

Final speed of the mobile = 5.2 m/s

Time, t = 8 s

We need to find the acceleration of the mobile. It can be given by the change in velocity divided by time. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(5.2-3.2)\ m/s}{8\ s}\\\\=0.25\ m/s^2$

So, the acceleration of the mobile is $0.25\ m/s^2$ .

3 0

3 years ago