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kompoz [17]
3 years ago
10

a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8

mm. the beam carries a uniform load of 6.5kN/m along the entire length and, in the same direction, a concentrated force of 4kN at the free end. (a) determine the max bending stress (b) determine the max transervse shear stress (c) determine the max shear stress in the beam
Engineering
1 answer:
djverab [1.8K]3 years ago
6 0

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

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6 0
1 year ago
The basic concept of feedback control is that an error must exist before some corrective action can be made?
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The correct answer is True.

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3 years ago
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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
IrinaVladis [17]

Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

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Change in diameter Δd = 0.04 mm

Final length l_{1} = 105.20 mm

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Shear modulus G = 25.4 G pa = 25.4 × 10^{3} M pa

We know that the relation between the shear modulus & elastic modulus is given by

G = \frac{E}{2(1 + \mu)}

25.5 = \frac{65.5}{2 (1 + \mu)}

\mu = 0.28

This is the value of possion's ratio.

We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

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\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

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l_{o} = 104.7 mm

This is the original length of the specimen.

5 0
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