How did dodge become a company
1 answer:
You might be interested in
Answer:
hello your question lacks some information attached is the complete question
A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in
(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in
B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi
(ii) Average shear stress at the web = 18.289 * 10^5 psi
(iii) Average shear stress at the Flange = 1.143 *10^5 psi
Explanation:
First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)
A) Calculate the maximum bending stress in tension and compression
lintel load = 10000 Ib
simple span = 6 ft
( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS
I = 53.54
i) The maximum bending stress (fb) in tension=
= =
= 0.287 * 10^6 Ib-in
ii) The maximum bending stress (fb) in compression
= =
= 0.7413*10^6 Ib-in
B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange
i) The average shear stress at the neutral axis
V = =
= 3.6*10^5 Ib
Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - ) *
= 5.878 in^3
t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi
ii) Average shear stress at the web ( value gotten from the shear stress at the flange )
t = 1.143 * 10^5 * (8 / 0.5 ) psi
= 18.289 * 10^5 psi
iii) Average shear stress at the Flange
t = VQ / Ib =
= 1.143 *10^5
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3