MATH7019: Test 2 Sample

April 11, 2011 in MATH7019

Test 2 on Thursday 19 April at 8.25 p.m

Please find a sample test here.

Note that the format will be the same as this.

Write down the bending moment for three loadings.
Exam Standard Beam I
Exam Standard Beam II
Binomial Distribution
Poisson Distribution

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April 19, 2012 at 2:28 pm

J.P. McCarthy

Q. 3

Using Macauley’s Method we have that:

$M(x)=-12[x-3]^2+R_Ax+M_A$ .

Now we solve

$EI\frac{d^2y}{dx^2}=-M$
$\Rightarrow EI\frac{d^2y}{dx^2}=12[x-3]^2-R_Ax-M_A$
$\Rightarrow EI\frac{dy}{dx}=12\frac{[x-3]^3}{3}-R_A\frac{x^2}{2}-M_Ax+C_1$
$\Rightarrow EI y(x)=4\frac{[x-3]^4}{4}-R_A\frac{x^3}{6}-M_A\frac{x^2}{2}+C_1x+C_2$ .

First using the fact that $y(0)=0$ :

$EI(0)=0=0-0-0+0+C_2\Rightarrow C_2=0$

Now as the ends are fixed, $y'(0)=0$ :

$EI(0)=0=0-0-0+C_1\Rightarrow C_1=0$

Now we look at $y(6)=0$ : the deflection at $x=6$ is zero:

$EIy(6)=EI(0)=0=$
$=3^4-R_A\frac{6^3}{6}-M_A\frac{6^2}{2}$

$\Rightarrow 81-36R_A-18M_A=0$
$\Rightarrow 2M_A+4R_A=9$ (A)

Now as the ends are fixed, $y'(6)=0$ :

$EIy'(6)=EI(0)=0$
$=4(3)^3-R_A\frac{6^2}{2}-M_A(6)$
$\Rightarrow 0=108-18R_A-6M_A$
$\Rightarrow 3R_A+M_A=18$ (B)
$\Rightarrow M_a=18-3R_A$ .

Put this into (A):

$2(18-3R_A)+4R_A=9$
$\Rightarrow 36-6R_A+4R_A=9$
$2R_A=27$
$R_A=+13.5$ kN

$\Rightarrow M_A=18-3(13.5)=-22.5$ kN m

$\Rightarrow EIy(x)=[x-3]^4-13.5\frac{x^3}{6}+22.5\frac{x^2}{2}$

$\Rightarrow y(x)=\frac{1}{EI}\left([x-3]^4-13.5\frac{x^3}{6}+22.5\frac{x^2}{2}\right)$

Any confusion about this not confirming to reality will result in the fact that I never explicitly said what orientation ( $+/-$ ) we were using. What we are certainly using are

$\frac{d^2M}{dx^2}=-w(x)$

and $EI\frac{d^2y}{dx^2}=-M(x)$ ,

although even then we could use

$EI\frac{d^2y}{dx^2}=M(x)$ …

You will be given the equation to use in the test and exam.

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MATH7019: Test 2 Sample

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