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VELOCITIES IN MECHANISMS THEORY OF MACHINE BY MA HELALY
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Introduction: Velocity
is important because: 1- It affects the time required to perform a given operation. 2- Power is the product of force and velocity. 3- Friction and wear on machine parts are also dependent on velocity. 4- Further, a determination of the velocities in a mechanism is required if an acceleration analysis is to be made.
Acceleration
is of interest because of its effect on inertia forces, which in turn influence stresses in the parts of a machine, bearing loads, vibration, and noise THEORY OF MACHINE BY MA HELALY
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VELOCITIES IN MECHANISMS BY METHOD OF
RELATIVE VELOCITIES THEORY OF MACHINE BY MA HELALY
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LINEAR VELOCITY Amplitude
Direction
V By scale from vector diagram Vc = 1.8
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OR
Vc = VB + VC/B then find D from triangle
Same triangle perp to original - opposite to link 2 = constant
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Example 3: In the toggle mechanism shown in Example 2, obtain the velocity of the ram F, The angular velocity of link EF. Solution: π π΄={ππππππ‘π’πππ2.ππ΄ π·πππππ‘πππβ₯ππ΄ π π΅=π π΄+π π΅π΄ π π΅={ππππππ‘π’πππ4.ππ΅ ?π·πππππ‘πππβ₯ππ΅ , π π΅π΄={ππππππ‘π’πππ3.π΄π΅ ?π·π ππππ‘πππβ₯π΄π΅ ππΆππ΅=ππππ π π·=π πΆ+π π·πΆ π π·={ππππππ‘π’πππ6.ππ· ?π·πππππ‘πππβ₯ππ· , π π·πΆ={ππππππ‘π’πππ5.πΆπ· ?π·π ππππ‘πππβ₯πΆπ· ΞΞ ππ·πΈ πππ π ππ πππ π ππππππ β΄ππ·π π=ππΈπ π=πΈπ·ππ π πΉ=π πΈ +π πΉπΈ
π πΉ={ππππππ‘π’ππ?π·πππππ‘πππβ , π πΉπΈ={ππππ ππ‘π’πππ7.πΉπΈ ?π·πππππ‘πππβ₯πΉπΈ THEORY OF MACHINE BY MA HELALY
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π7=ππβπ£ππ.π πππππΈπΉ=12.57 ππππ π.π€., ππΉ=ππβ π£ππ.π ππππ=123.31 πππ β
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Example 4: For the Whitworth mechanism shown in Example 1, determine the absolute velocity of the tool support C; also find the angular velocities of links QB and BC. Solution: Consider point T on link 4 under point A which on the slider 3. VA = 900 cm/s (β΄ OA) π π=π π΄+π ππ΄ π π={ππππππ‘π’πππ4.ππ ?π·πππππ‘πππβ₯ππ , π π π΄={ππππππ‘π’ππ?π·πππππ‘πππβ₯ππ(ππππ 4) ππππ΅ =ππ‘ππ π πΆ=π π΅+π πΆπ΅ π πΆ={ππππππ‘π’ππ?π·πππππ‘πππβ , π πΆπ΅={πππ πππ‘π’πππ5.π΅πΆ ?π·πππππ‘πππβ₯π΅πΆ From the velocity diagram: π4=ππ‘βπ£ππ.π πππππ π=23.81 ππππ π.π€., π5=ππβπ£ππ.π πππππ΅πΆ=4.65 ππππ π.π.π€. ππΆ=ππβπ£ππ.π ππππ=3.44 ππ β
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VELOCITIES IN MECHANISMS BY METHOD OF
instant Centers THEORY OF MACHINE BY MA HELALY
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Instantaneous center of a velocity: When two links (bodies), either both moving or one moving and one fixed, the instantaneous center is: a- a point in both bodies, b- a point at which the two bodies have no relative velocity (the point has the same velocity in each body), c- A point about which one body may be considered to rotate relative to the other at a given instant. Notes: - When the two links are directly connected together, the center of the connecting joint is an instantaneous center for the two links, - If not directly connected, an instantaneous center exists for a given phase of the linkage.
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Instantaneous Center Notation: The system of labeling instantaneous centers is - The instantaneous center for link (2) relative to link (1) is labeled 21. Link(1) has the same instantaneous center relative to link(2), when link(2) is considered the fixed link, link(1) appears to be rotating in the opposite sense (Ο12 = - Ο21) relative to link(2). - Since points 21 and 12 are the same point, either designation is acceptable, 12 is preferred.
-Pin connection: each pin connection is an instant center (12, 14 remain fixed, they are called fixed centers), (23, 34 are called moving centers, they move relative to the frame)
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Number of instantaneous centers for a mechanisms:
Any two links in a mechanism have motion relative to one another and have a common instantaneous center. The number of instant centers is equal to all possible combinations of two from the total numbers of links. Let, n = number of links. Then the number of instant centers is:
π=π(πβ1) / 2! THEORY OF MACHINE BY MA HELALY
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e- Primary instantaneous centers: All instant centers which can be found by inspection are called primary instant centers, and then we can locate the remaining by applying Kennedy's theorem. Primary instant centers can be summarized as: 1- Instant center for pin connecting links. 2- Instant center for a sliding body. 3- Instant center for a rolling body. 4- Direct contact mechanisms: a- For sliding contact: intersection of the common normal and the line of centers. b- For rolling contact: at the point of contact. f- Circle diagram method for locating instantaneous centers: - All primary centers must be located first, - Points are laid out along a circle, each point represent a link, - All possible straight lines joining these points represent the instant centers, - First, all centers which have located are drawn in as solid lines, - The remaining are represented by dotted lines, - In order to locate these centers, find any two triangles which a dotted line completes, These two triangles represent two lines their intersection is the required instantaneous center, - After an instant center has been located, it is drawn in as a solid line on the circle diagram. (The two triangles must have a common side which is dotted). THEORY OF MACHINE BY MA HELALY
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Locating instantaneous centers in a four-bar mechanism THEORY OF MACHINE BY MA HELALY
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Locating instant centers in the slider-crank linkage THEORY OF MACHINE BY MA HELALY
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Velocity analysis using instantaneous centers: THEORY OF MACHINE BY MA HELALY
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Linear velocities by instantaneous centers: - The magnitude is equal to the product of the radius of rotation (distance between the point and the instant center 1i). -It must be β΄ the radius of rotation of the point.
Angular velocities by instantaneous centers:
From the last equation we can conclude that the angular-velocity ratio for any two links in a mechanism is inversely as the distances from the instant centers in the frame about which the links are rotating to the instant center which is common to the two links. THEORY OF MACHINE BY MA HELALY
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If the common instant center ij lies in-between 1i & 1j, then Οi and Οj are in opposite direction, and if not, then Οi and Οj are in the same direction. THEORY OF MACHINE BY MA HELALY
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Example 1: For the Whitworth mechanism shown, determine the absolute velocity VC of the tool support, when the driving link OA rotates at a speed such that VA = 900 cm/s, as shown, also find the angular velocities of links QB and BC. OQ = 135, OA = 270, QB = 160, and BC = 550 mm.
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