Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:

Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:

Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:

Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
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
100: D, third law of motion
101: D, second law of motion
Answer:
D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted
Explanation:
access-list 90
deny 10.168.7.0 0.0.0.255
permit 10.168.7.10
permit 10.168.7.11
permit 10.168.7.12
deny any