Please answer questions 4-7 ONLY. Make sure your answers are clear and you show

Question

Please answer questions 4-7 ONLY. Make sure your answers are clear and you show the steps to your solution for best answer. Thank you.

The figure defines the following vectors: A is the crank (length L1) B is the connecting rod or coupler (length L2) C is the "fixed" link (length X(t)). In addition: S = X - L2 is the slider (or piston ) stroke or displacement theta 1(t) is the crank angle write the kutzbach criterion, identifying the terms, and predict the mobility, m, of this mechanism. What is the magnitude of total stroke |S| of the slider in terms of the constant parameters? Comment on any necessary practical aspects of the ratio L2/L1 (for example, what if L2 is shorter than L1) and consider this for various crank angles. What does the Grahof's law require? Write the kinematic loop closure equation (a) in terms of the vectors A, B & C and (b) in terms of L1, L2, X, theta 1 & theta 2 (note 2 eqns are required). Assuming theta 1 is known at some instant, identify the known and unknown magnitudes and directions of A, B & C and solve for the unknowns in terms of the given parameters using (a) trigonometry and (b) the chace solutions. Derive an expression for S in terms of L1, L2 and 01. eliminating theta 2. Assuming X (or S) is known at some instant, identify the known and unknown magnitudes and directions of A, B & C and solve for the unknowns in terms of the given parameters using (a) trigonometry and (b) the chace solutions. Derive an expression for dS/dt (the slider vel) in terms of L1, L2 and theta 1.

Explanation / Answer

4.

a)

Vector equation for loop closure would be A + B = C

b)

In terms of theta1, theta2, L1, L2, and x we can write as follows:

Horiontal component of A = L1*cos theta1

Horizontal component of B = L2*cos theta2

Thus, L1*cos theta1 + L2*cos theta2 = x...................1

Vertical component of A = L1*sin theta1

Vertical component of B = L2*sin theta2

We have, L1*sin theta1 = L2*sin theta2.......................2

5.

Magnitude of A = L1 (known)

Direction of A = theta1 (known)

Magitude of B = L2 (known)

Direction of B = theta2 (unknown)

Magnitude of C = x (unknown)

Direction of C = horizontal (known)

a)

We have theta2 and x as unknown.

From eqn 2, we get theta2 = asin [(L1 / L2)*sin theta1]

cos theta2 = sqrt (1 - sin^2 theta2)

cos theta2 = sqrt (1 - (L1 * sin theta1 / L2)^2)

cos theta2 = sqrt (L2 ^2 - L1 ^2 *sin^2 theta1) / L2

L2*cos theta2 = sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)...................3

Now putting it in eqn 1, we get

x = L1*cos theta1 + sqrt (L2 ^2 - L1 ^2 *sin^2 theta1).............4

b)

Chace solution:

Draw x-y axis and an origin.

Draw a line of length L1 at an angle theta1 above x-axis. This denotes vector A.

From the other end of this line, draw an arc of radius = L2. Mark the point where the arc crosses the x-axis. This point would be the slider position.

draw a line from arc's centre to slider position. This denotes vector B.

Join origin and end of vector B. This becomes vector C.

Now measure theta2, angle of vector B with horizontal and measure length of vector C which is nothing but x.

6.

Magnitude of A = L1 (known)

Direction of A = theta1 (unknown)

Magitude of B = L2 (known)

Direction of B = theta2 (unknown)

Magnitude of C = x (known)

Direction of C = horizontal (known)

Now unknowns are theta1 and theta2.

a)

From eqn 2, we get theta2 = asin [(L1 / L2)*sin theta1]

From eqn 1 and 3 we get,

x = L1*cos theta1 + sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)

Thus, L1*cos theta1 = x - sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)

cos theta1 = [x - sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)] / L1

theta1 = sqrt [ (x - sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)) / L1]

b)

Chace solution:

Draw origin and x-y axis.

On x-axis mark distance x from origin. This would be vector C.

From origin, draw an arc of radius L1.

From end of vector C draw an arc of length L2.

Note the point where both these arcs intersect.

Make a straight line from origin to intersection point. This would be vector A. Meausure its angle theta1 from x-axis.

Make a straight line from intersection point to end of vector C. This would be vector B. Measure its angle theta2 from horizontal.

7.

From eqn 4,

x = L1*cos theta1 + sqrt (L2 ^2 - L1 ^2 *sin^2 theta1)

velocity v = dx / dt

Taking derivative,

v = L1*(-sin theta1)*(d theta1 / dt) + (1/2)*[(L2 ^2 - L1 ^2 *sin^2 theta1)^ (-1/2)] *(-L1 ^2)*(2*sin theta1 *cos theta1)

v = -L1*sin theta1 * (d theta1 / dt) - L1 ^2 * sin theta1 * cos theta1 *(L2 ^2 - L1 ^2 *sin^2 theta1)^ (-1/2)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---