The statically-indeterminate composite beam AB of length 2L is fixed at A (x=0) and is simply supported at B (x=2L). The beam is composed of a core of modulus E0 of constant square cross section of dimensions 2h0×2h0 bonded inside a sleeve of modulus 2E0 and constant square cross section of outer dimensions 4h0×4h0, as indicated in the figure. The beam is loaded by a downward concentrated load of magnitude P applied at the midspan of the beam (x=L) as indicated.
The KNOWN quantities in this problem are P[N], L[m], h0[m], and E0[Pa]. In symbolic expressions, enter P, L, h0 and E0 as P, L, h_0 and E_0, respectively.
1)Consider the SD companion problem obtained by selecting the roller at B as the redundant support, and obtain the bending moment resultant M(x) in terms of x, L, the unknown redundant reaction RBy (enter this as R_y^B), and P.
for 0≤x≤L,M(x)=
for L≤x≤2L,M(x)=
2)Obtain a symbolic expression for the redundant reaction RBy in terms of P.
RBy=
3)Obtain symbolic expressions for the curvature of the beam at the fixed support A, (1ρ(x=0)), and at the midspan of the beam, (1ρ(x=L)), in terms of P, L, h0, and E0.
1ρ(x=0)=
1ρ(x=L)=
4)Obtain symbolic expressions in terms of the known quantities for the maximum tensile stresses in the core and in the sleeve on the x=0 cross section, and indicate at what y each of them occurs. Express your answers in terms of P, L, and h0.
σmaxcore=
at y= -
σmaxsleeve=
at y=
Did anyone found a full solution to this?
I found:
For 1):
for 0≤x≤L,M(x)=P*(x-L) +(2*L -x) * R_y^B
for L≤x≤2L,M(x)= R_y^B *( 2*L- x)
For 2)
RBy= P/4, BUT THIS IS INCORRECT !
And I am stuck here! Which I think is weird, My Matlab graphs show everything to conform with RBy= P/4
UF did you figure out any more of the second problem(other parts). I'm majorly stuck here
i found Q2_2_4 the two y's.the first one is h_0 and the second one is 2*h_0
Thanks guys! Anything u have on the the other problem helps too. I'm looking for fill. I need a 70 to pass class and am close but not there.....this is the first problem:
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Homework Help: Physics
Posted by GaryG on Monday, August 25, 2014 at 11:47am.
CONICAL SHAFT WITH DISTRIBUTED TORSIONAL LOAD
The conical shaft AB is made of steel, with shear modulus G0. The shaft has length L and is free at A (where x=0) and fixed at B (where x=L). The shaft is solid, with a linear taper so that the radius of the generic (circular) cross section is defined by the equation R(x)=R0(xL), where R0 is the radius at the wall, B. The shaft is subjected to an applied distributed torque per unit length tx(x)=t0(xL)4, where t0 is the magnitude of the applied distributed torque at B (x=L).
The KNOWN quantities in this problem are:
L=2m
G0=70GPa
R0=2cm
t0=2kN⋅m/m
In symbolic expressions, do NOT substitute ANY of the given numerical values of the known quantities, enter L, G0, R0, and t0 as L, G_0, R_0, and t_0, respectively, and enter π as pi.
1)Obtain a symbolic expression for the axial torque resultant in the bar T(x) in terms of x, t0, and L.
T(x)=
2)Compute the numerical value of the reaction TBx at support B, in units of kN·m. (Note: You will need to provide your answer to the second decimal digit.)
TBx=
3)Obtain a symbolic expression for the twist rate along the shaft, dφdx, in terms of x, G0, t0, and R0.
dφdx(x)=
4)Compute the numerical magnitude (absolute value) of the maximum shear stress in the shaft, τmax, in units of MPa:
τmax=
Enter symbolic expressions for the location (r,x) at which this maximum shear stress occurs, in terms of R0 and L:
r=
x=
5)Obtain a symbolic expression for the angle of rotation of the free end A of the shaft in terms of t0, G0, R0, and L:
φ(x=0)=
Compute the numerical value of the angle of rotation of the free end, in units of radians. (Note: You will need to provide your answer to the third decimal digit.):
φ(x=0)=
To solve this problem, we need to apply the principles of mechanics and statics to determine the bending moment, redundant reaction, curvature, and maximum tensile stresses in the composite beam. Here's how we can approach each of the questions.
1) To obtain the bending moment resultant M(x) in terms of x, L, RBy, and P, we need to consider two regions: 0 ≤ x ≤ L and L ≤ x ≤ 2L.
For 0 ≤ x ≤ L, the beam is loaded by a downward concentrated load of magnitude P at the midspan (x = L). The bending moment M(x) in this region can be determined using the equation for a simply supported beam with a concentrated load:
M(x) = -Px/2
For L ≤ x ≤ 2L, the beam is a statically indeterminate structure with the redundant reaction RBy at the roller support B. The bending moment M(x) in this region can be found using the equation for a simply supported beam:
M(x) = -P(L - x)/2 - RBy(x - L)
2) To obtain the expression for the redundant reaction RBy in terms of P, we need to consider the equilibrium of forces in the vertical direction at B. Since the beam is in equilibrium, the sum of vertical forces must be zero. Considering the downward load P and the upward reaction RBy, we have:
P = RBy
Therefore, the expression for RBy is simply:
RBy = P
3) To determine the curvature at the fixed support A (x = 0) and at the midspan of the beam (x = L), we need to use the equation for curvature in terms of bending moment, modulus of elasticity, and moment of inertia.
The curvature (1/ρ) at x = 0 can be found using the equation:
(1/ρ)(x=0) = M(x=0) / (E0 * I)
Where I is the moment of inertia of the composite beam, which is the sum of the moments of inertia of the core and the sleeve. The moment of inertia for a square cross-section is (1/12) * w * h^3, where w and h are the width and height of the cross-section.
The curvature (1/ρ) at x = L can be found using the equation:
(1/ρ)(x=L) = M(x=L) / (2E0 * I)
Note that the modulus of elasticity for the sleeve is given as 2E0.
4) To obtain the maximum tensile stresses in the core and sleeve at the x = 0 cross-section and the corresponding y coordinates, we can use the formula for stress:
σmax = M(x=0) * y / I
For the core, the maximum tensile stress occurs along the bottom edge of the cross-section with y coordinate equal to (-h0/2).
For the sleeve, the maximum tensile stress occurs along the bottom edge of the cross-section with y coordinate equal to (-2h0/2).
Substituting the expressions for M(x=0) and I for each region, we can calculate the maximum tensile stresses in the core and sleeve.
Note: It's important to carefully apply the appropriate formulas and equations based on the given information and assumptions.