A pre-stressed concrete beam 300mm wide and 600mm deep is pre-stressed using 5 high tension bars of 12mm f provided at 220mm from the soff it of the beam. Effective stress in the steel is 800N/mm2. The bending moment that must be applied at the section to just avoid tension at the soff it of the beam will be __________ KNm.
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Solution
Area = 300 × 600 = 18 × 104 mm2
= 452389 Ne = 300 – 200 = 80mm
\(Z=\frac{bd^{2}}{6}=18\times 10^{6}mm^{3}\)
so,P⁄A+Pe⁄Z-M⁄Z=0
\(\frac{452389}{18\times 10^{4}}+\frac{452389\times 80}{18\times 10^{6}}=\frac{M}{18\times 10^{6}}\)
M = 81.36 KNm
Elements of stiffness matrix K11 and K22 are
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Solution
For K11
For K11=\(\frac{4EI}{L}+\frac{4EI}{L}+\frac{4EI}{L}=\frac{12EI}{L}\)
For K22
K22=\(\frac{4EI}{L}\)
Shape of influence line diagram for maximum bending moment in respect of simply supported beam.
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Solution
In simply supported beam, maximum bending moment will occur below the load it self
So max B.M =x⁄ι(ι-x)
An overhang beam of uniform EI is loaded as shown in the above figure. The deflection at free end will be
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Solution
θ=\(\frac{M(2L)}{3EI}\)
θ=\(\frac{PL(2L)}{3EI}\)
δ2=θ×L=\(\frac{2}{3}\frac{PL^{3}}{EI}\)
A=δ1.δ2=\(\frac{PL^{3}}{3EI}+\frac{2}{3}\frac{PL^{3}}{EI}=\frac{PL^{3}}{EI}\)
A 100N force applied on a bent handle as shown in the figure. The moment about base O will be __________ Nm.
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Solution
Taking summation of moments about point o
M = 100 cos 15°(0.5 sin 30°+ 1 sin 60°) – 10° sin 15° (0.5 cos 30 + 1 cos 60°)
= 96.592 (0.25 + 0.866) – 25.88 (0.433 + 0.5)
= 107.79 – 24.146
= 83.64 Nm
As x varies from –1 to + 3, which one of the following describes the behaviour of the function f (x) = x3 – 3x2 + 1?
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Solution
f(–1) = (– 1)3– 3(–1)2 + 1 = – 1 – 3 + 1
f(0) = (0)3– 3(0)2+ 1 = 1
f(1) = (1)3– 3(1)2 + 1 = – 1
f(2) = (2)3– 3(2)2+ 1 = 8– 12 + 1 = – 3
f(3) = (3)3– 3(3)2+ 1 = 27 – 27 + 1= 1
For –1 to 0:
f is increasing:
For 0 to 2:
f is decreasing For 2 to 3:
f is increasing.
The value of x for which the matrix A=\(\begin{bmatrix} 3 & 2 & 4 \\ 9 & 7 & 13 \\ -6 & -4 & -9+x \end{bmatrix}\)
has zero as an eigenvalue is ________
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Solution
1
For eigen value of A is to be zero, det (A) =0
i.e.\(\begin{bmatrix} 3 \, 2 \, 4 & & \\ 9 \, 7 \, 13 & & \\ -6 \, -4 \, -9+x & & \end{bmatrix}\)=0
⇒3[7 (– 9 + x) + 4 (13)] –2 [9(–9 + x) + 6(13)]+ 4[9(– 4) + 6(7)] =0
⇒3[– 63 + 7x + 52] – 2[–81 + 9x + 78] + 4[–36 + 42] = 0
⇒3[– 63 + 7x + 52] – 2[–81 + 9x + 78] + 4[–36 + 42] = 0
⇒3[– 63 + 7x + 52] – 2[–81 + 9x + 78] + 4[–36 + 42] = 0
⇒3x – 3 = 0 ⇒ x = 1
Given the following statements about a function f: R→R,select the right option:
P: If f(x) is continuous at x= x0, then it is also differentiable at x= x0.
Q: If f(x) is continuous at x=x0, then it may not be differentiable at x= x0.
R: If f(x) is differentiable at x= x0, then it is also continuous at x= x0.
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Solution
Since continuous function is always differential and differential function is not necessary be a continuous function (i.e., it may or may not be continuous function).
The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is _______
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Solution
1
Let first moment be x
Now,second moment is x2 + x
According to question
⇒x2+ x=2
⇒x2+ x – 2 = 0
⇒x2+ 2x – x –2 =0
⇒x (x +2) –1(x +2)= 0
⇒x =1, – 2 (Negative value rejected)
⇒x = 1
But in Poisson distribution
Mean = First moment (x)
∴Mean= x = 1
Let M4= 1, (where I denotes the identity matrix) and M≠1,M2≠1 and M3≠1. Then, for any natural number k, M-1 equals:
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Solution
M4=I...(i)
⇒(M4)k= (I)k
⇒M4K = I (∵I.I = I)...(ii)
From (i) and (ii), we get
M4. M4K= I.I
⇒M4(K+1)= I(∵ I.I = I)
⇒M4(K+1). M–1= M–1
⇒M4K + 4–1 = M–1
M-1 = M4K+3