A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a
nd width b = 12 mm (0.47 in.); the distance between support points is 43 mm (1.69 in.).(a) Compute the flexural strength if the load at fracture is 292 N (66 lbf). (b) The point of maximum deflection, ?y, occurs at the center of the specimen and is described by where E is the modulus of elasticity and I is the cross-sectional moment of inertia. Compute ?y at a load of 266 N (60 lbf). Assume that the elastic modulus for the glass is 60 GPa. What is the flexural strength of this sample in MPa if the load in part (a) is applied? What is I, the moment of inertia, in m4? What is the deflection, ?y, in mm, at the load given in the problem?
1 answer:
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Answer:
where and
be the inner radius, outer radius of the annalus.
Explanation:
Let ,
and
be the inner radius, outer radius and length of the given annulus.
Temperatures at the inner surface, and at the outer surface,
.
Let q be the rate of heat transfer at the steady-state.
Given that, the heat flux at r=3cm=0.03m is
.
This heat transfer is same for any radial position in the annalus.
Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.
Now, according to Fourier's law:
where,
K=Thermal conductivity of the material.
T= temperature at any radial distance r.
A=Area through which heat transfer is taking place.
Here,
Variation of temperature w.r.t the radius of the annalus is
Putting the values from the equations (ii) and (iii) in the equation (i), we have
[as
, and
]
This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.
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 = =
= 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
