Ask question

Ask question

All categories

3 years ago

8

The W16x50, steel beam below has a span of 26' and is subjected to a 2.3 k/ft uniform distributed loading. If a 8 kip load is al

so applied at the mid point, what is the maximum bending stress in the beam

1 answer:

MariettaO [177]3 years ago

5 0

Answer: hello the steel beam related to your question is missing attached below is the missing diagram

answer : max stress = -14.32 ksi

Explanation:

<em>For a W16 * 50 beam </em>

I = 659 in

d = 16.26 in

For steel : E = 29 * 10^3 ksi

Span ( length ) of steel beam = 26'

uniformly distributed load on the steel beam = 2.3 k/ft

load at middle = 8 kip , Distance from middle to either Ra or Rb = 13'

<em>First step : calculate the value of Rb and Ra </em>

∑Ma = 0

0 = ( 8 * 13 )+ (26 * 2.3 * 13 ) - Rb * 26

∴ 25 Rb = 881.4

Hence: Rb = 35.256 kip

∑fy = 0

0 = Ra + Rb - 2.3 * 26 - 8

Ra = 67.8 - Rb = 67.8 - 35.256

= 32.54 kip

second step :

Mmax = Mmidpoint

∴ Mmax = Ra * 13 - 2.3 ( 13 ) * (26/4 )

= 32.54 * 3 - 2.3 ( 13 ) * (26/4 )

= 97.62 - 194.35 = -96.73 kip fit

<u>Finally determine the maximum bending stress </u>

бmax = M * ( d/2 ) / I

= - 96.73 * ( 16.26 / 2 )*12 / 659

= -96.73 * 97.56 / 659 = -14.32 ksi

<u />

<u />

You might be interested in

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago

State the number of terms for each of the following algebraic expression 2x+1

harina [27]

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

Hi please follow me also if you can and thanks.

6 0

3 years ago

A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?

Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is $L=100\ cm$

Strain experienced is $\epsilon=82\%=0.82$

Strain is the ratio of the change in length to the original length

$\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm$

Therefore, new length is given by (Considering the load is tensile in nature)

$\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm$

Thus, option (b) is correct.

8 0

3 years ago

A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055

Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

$K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}$

So the mean conductivity

$K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )$

$T_m=\dfrac{160+273+40+273}{2}$

$T_m=373 K$

$K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )$

$K_m=0.112 \frac{W}{mK}$

So heat conduction through cylinder

$Q=kA\dfrac{\Delta T}{L}$

$Q=0.112\times \pi \times 0.15^2\times 120$

Q=0.95 W/m

4 0

3 years ago

An engineering drawing shows the: (A) dimensions, tolerances, cost, and sales or use volume of a component.(B) dimensions, toler

Leto [7]

Answer:

(B) dimensions, tolerances, materials, and finishes of a component.

Explanation:

An engineering drawing :

An engineering drawing is a technical drawing which draws the actual component .

An engineering drawing shows

1. Materials

2.Dimensions

3.Tolerance

4.Finishes of a component

Engineering drawing does not shows any information about the cost of component.

So the option B is correct.

3 0

3 years ago