A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe
1 answer:
Answer:
moment of inertia = 4.662 * 10^6
Explanation:
Given data :
Mass of machine = 400 kg = 400 * 9.81 = 3924 N
length of span = 3.2 m
E = 200 * 10^9 N/m^2
frequency = 9.3 Hz
Wm ( angular frequency ) = 2 = 58.434 rad/secs
also Wm = ------- EQUATION 1
g = 9.81
deflection of simply supported beam
t =
insert the value of t into equation 1
W =
make I the subject of the equation
I ( Moment of inertia about the neutral axis ) =
I = = 4.662 * 10^6
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
