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skelet666 [1.2K]

2 years ago

13

A cable consists of a steel wire 2.5mm in diameter, surrounded by six bronze wires of the same size. If the allowable stress in

the bronze is 60MPa, find the safe load for the cable and the extension under this load on the length of 30m (Est= 200GPa and Ebr = 80GPa).

1 answer:

DENIUS [597]2 years ago

6 0

Answer:

Explanation:

given data:

steel diamter = 25 mm

allowable stress in bronze = 60 MPa

length of wire = 30 m

Ebr=80GPa

Est = 200GPA

AREA of wire $= \frac{\pi}{4} 2.5^2 = 4.906 mm2$

total load = 6*Pbr + Pst .............1

where

Pbr is load due to bronze

Psr is load due to steel

elongationis given as

$\frac{Pbr*l}{Ebr*A} =\frac{Pst*l}{Est*A}$

$Pbr = \frac{ Ebr}{Est} Pst$

$= \frac{ 80}{200}Pst$

force on bronze wire

$\sigma br = \frac{Pbr}{A}$

$60 = \frac{Pbr}{4.90625}$

Pbr = 294.375 N

from eqt 2

$294.375 = \frac{80}{200} Pst$

Pst = 735.93 N

SAFE LOAD from eq 1

P = 6 *294.375 + 735.937 = 2502.18 N

EXtension under load

$\delta t = \frac{Pbr L}{AEbr} = \frac{294.37*30*10^3}{4.90625*80*10^3}$

$\delta t = 22.5 mm$

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