All categories

4 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.

Engineering

1 answer:

Novosadov [1.4K]4 years ago

5 0

Answer:

i) SF: $v(x) = \frac{(w_0* x )^2}{2L}$

ii) BM : $= \frac{(w_0*x)^3}{6L}$

Explanation:

Let's take,

$\frac{y}{w_0} = \frac{x}{L}$

Making y the subject of formula, we have :

$y = \frac{x}{L} * w_0$

For shear force (SF), we have:

This is the area of the diagram.

$v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0$

$= \frac{(w_0* x )^2}{2L}$

The shear force equation =

$v(x) = \frac{(w_0* x )^2}{2L}$

For bending moment (BM):

$BM = v(x) * \frac{x}{3}$

$= \frac{(w_0* x )^2}{2L} * \frac{x}{3}$

$= \frac{(w_0*x)^3}{6L}$

The bending moment equation =

$= \frac{(w_0*x)^3}{6L}$

You might be interested in

You wish to design a cantilever beam with a square cross section and length L that can support an end load of F without yielding

koban [17]

Answer:

with a square cross section and length L that can support an end load of F without yielding. You also wish to minimize the amount the beam deflects under load. What is the free variable(s) (other than the material) for this design problem?

a. End load, F.

b. Length, L.

c. Beam thickness, b

d. Deflection, δ

e. Answers b and c.

f. All of the above.

8 0

3 years ago

The distillation column in Figure 3 is set up for so-called boil-up (V) control. It has

11111nata11111 [884]

A distillation column is an essential item used in the distillation of liquid mixtures to separate the mixture into its component parts, or fractions, based on the differences in volatilities. Fractionating columns are used in small scale laboratory distillations as well as large scale industrial distillations.

3 0

3 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 years ago

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

$W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ$

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0

3 years ago

Select the correct answer.

Rina8888 [55]

A: Inexhaustible resources are limited

is true

6 0

3 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.​

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.