Page 205 - Shigley's Mechanical Engineering Design

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180 Mechanical Engineering Design
3 The upward deflection of point A is (Fl/AE) AD − Np, where the first term is the
elastic stretch of AD, N is the number of turns of the nut, and p is the pitch of the
thread. Thus, the deflection of A in meters is
F A (0.6)
y A = − (1)(0.0015)
π 2 9
(0.010) (200)(10 ) (4)
4
−3
−8
= 3.8197(10 )F A − 1.5(10 )
The upward deflection of point C is (Fl/AE) CE , or
F C (0.8) −8
y C = = 5.093(10 )F C (5)
π 2 9
(0.010) (200)(10 )
4
Equations (4) and (5) will now serve as the boundary conditions for Eq. (3). At
4
x = 0, y = y A . Substituting Eq. (4) into (3) with x = 0 and EI = 1.25(10 ), noting
that the singularity function is zero for x = 0,gives
−4
−4.7746(10 )F A + C 2 =−18.75 (6)
At x = 0.2 m, y = 0, and Eq. (3) yields

−3
−1.3333(10 )F A + 0.2C 1 + C 2 = 0 (7)
At x = 0.35 m, y = y C . Substituting Eq. (5) into (3) with x = 0.35 m and EI =
4
1.25(10 ) gives
−3
−4
−4
−7.1458(10 )F A + 5.625(10 )F B − 6.3662(10 )F C + 0.35C 1 + C 2 = 0 (8)
Equations (1), (2), (6), (7), and (8) are five equations in F A , F B , F C , C 1 , and C 2 .
Written in matrix form, they are
⎧ ⎫ ⎧ ⎫
−1 1 −1 0 0 0
⎡ ⎤
⎪ F A ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪
4 0 −3 0 ⎪ ⎪ ⎪ 0 ⎪
⎪
⎪
⎪
⎪
⎢ 0 ⎥ ⎨ F B ⎬ ⎨ ⎬
−4
⎢ −4.7746(10 ) 0 0 0 1 ⎥ F C = −18.75
⎢
⎥
−1.3333(10 ) 0 0 0.2 1 ⎪ ⎪ ⎪ 0 ⎪
⎣ −3 ⎦ ⎪ ⎪ ⎪ ⎪
⎪ C 1 ⎪ ⎪ ⎪
−4
−3
−4
⎪
⎪
⎪
⎪
−7.1458(10 ) 5.625(10 ) −6.3662(10 ) 0.35 1 ⎩ C 2 ⎭ ⎩ 0 ⎭
Solving these equations yields
Answer F A = 2988 N F B = 6971 N F C = 3983 N
C 1 = 106.54 N · m 2 C 2 =−17.324 N · m 3
Equation (3) can be reduced to
3 3 −3
y =−(39.84x − 92.95 x − 0.2 − 8.523x + 1.386)(10 )
Answer At x = 0, y = y A =−1.386(10 ) m =−1.386 mm.
−3
Answer At x = 0.35 m, y = y C =−[39.84(0.35) − 92.95(0.35 − 0.2) − 8.523(0.35)
3
3
−3
−3
+ 1.386](10 ) = 0.203(10 ) m = 0.203 mm