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2 years ago

. Ropes made from are typically very weak.

Engineering

1 answer:

Umnica [9.8K]2 years ago

6 0

Answer: cotton and/or known as Cotton rope

Explanation: It is a very weak fiber that has less strength than cotton. So its typically very weak

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Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit cont

Amiraneli [1.4K]

Answer:

Explanation:

6 0

3 years ago

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft.

saul85 [17]

Answer:

a. $L_o$ = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, $L_o$ = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

$L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)$

For the school, $K_{LL}$ = 2

Therefore, we have;

$L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf$

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, $W_d$ = b × $W_{D + L}$ =

∴ $W_d$ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, $W_d$ = 812.8 ft./lb

d. For the uniformly distributed load, we have;

$V_{max}$ = 812.8 × 26/2 = 10566.4 lbs

$M_{max}$ = 812.8 × 26²/8 = 68,681.6 ft-lbs

$v_{max}$ = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

$P_L$ = 1000 lb

$W_{D}$ = 20 × 16 = 320 lb/ft.

$V_{max}$ = 1,000 + 320 × 26/2 = 5,160 lbs

$M_{max}$ = 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

$v_{max}$ = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

7 0

3 years ago

What is a jack plane used for

kolezko [41]

A jack plane is a general purpose woodworking bench plane, used for dressing timber down to the correct size in preparation for truing and/or edge jointing.

I hope this helps you :)

4 0

3 years ago

Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is

frozen [14]

Answer:

$x = 0.944$

Explanation:

Steam at outlet is an superheated steam, since $T > T_{sat}$ . From steam tables, the specific enthalpy is:

$h_{out}=2682.4\,\frac{kJ}{kg}$

The throttle valve is modelled after the First Law of Thermodynamics:

$h_{in} = h_{out}$

Hence, specific enthalpy at inlet is:

$h_{in}=2682.4\,\frac{kJ}{kg}$

The quality in the steam line is:

$x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }$

$x = 0.944$

6 0

3 years ago

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

3 years ago