The critical loads of thin columns depend on the end conditions of the column. The value of the Euler load P1 in Example 4 was derived under the assumption that the column was hinged at both ends. Suppose that a thin vertical homogeneous column is embedded at its base (x = 0) and free at its top (x = L) and that a constant axial load P is applied to its free end. This load either causes a small deflection δ as shown in Figure 5.2.9 or does not cause such a deflection. In either case the differential equation for the deflection y(x) is
Figure 5.2.9 Deflection of vertical column in Problem 24
- (a) What is the predicted deflection when δ = 0?
- (b) When δ ≠ 0, show that the Euler load for this column is one-fourth of the Euler load for the hinged column in Example 4.
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Chapter 5 Solutions
First Course in Differential Equations (Instructor's)
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The root can be located with false position method.arrow_forwardWhich of the following statements is true for regular, isotropic beams: Maximum bending or flexural stress occurs at the point in a cross-section that is furthest from the neutral axis. Flexural stress varies linearly along the cross-section of the beam. (A) B) (D) (E) F Maximum horizontal shearing stress occurs at the neutral axis. Horizontal shearing stress varies "parabolically" along the cross-section of the beam. Maximum bending or flexural stress occurs at the neutral axis. Flexural stress varies linearly along the cross-section of the beam. Maximum horizontal shearing stress occurs at the neutral axis. Horizontal shearing stress varies "parabolically" along the cross-section of the beam. Maximum bending or flexural stress occurs at the neutral axis. Flexural stress varies linearly along the cross-section of the beam. Maximum horizontal shearing stress occurs at the point in a cross-section that is furthest from the neutral axis. Horizontal shearing stress varies "parabolically"…arrow_forwardIt may surprise you to learn that the collision of baseball and bat lasts only about a thou- sandth of a second. Here we calculate the average force on the bat during this collision by first computing the change in the ball's momentum. The momentum p of an object is the product of its mass m and its velocity v, that is, p = mv. Suppose an object, moving along a straight line, is acted on by a force F = F(t) that is a continuous function of time. (a) Show that the change in momentum over a time interval [to, t1] is equal to the integral of F from to to t1; that is, show that p(t) – p(to) = |" F(t) dt This integral is called the impulse of the force over the time interval. (b) A pitcher throws a 90-mi/h fastball to a batter, who hits a line drive directly back to the pitcher. 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According to Newton's Second Law of Motion, there are two forcës acting on the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa) as shown in Figure 1. Fa = -cv ITM EUTM FUTM * UTM TM Fg= -mg x(t) UTM UT UTM /IM LTM UTM UTM TUIM UTM F UT GROUND Figure 1: Force acting on body of free-fall where x(t) is the position of the parachutist from the ground at given time, t is the time of fall calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration, v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the position is given by the equations below: EUTM PUT v(t) = mg -et/m – 1) (Eq. 1.1) x(t) = x(0) – Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the critical position to deploy the parachutes is at 762 m from the ground…arrow_forwardObtain characteristics for y2 Uxx + Uyy = %3Darrow_forward1. Suppose that a car weighing 4000 pounds is supported by four shock absorbers Each shock absorber has a spring constant of 6500 lbs/foot, so the effective spring constant for the system of 4 shock absorbers is 26000 lbs/foot. 1. Assume no damping and determine the period of oscillation of the vertical motion of the car. Hint: g=32 ft/sec². T= 0.436 seconds. 2. After 10 seconds the car body is 1 foot above its equilibrium position and at the high point in its cycle. What were the initial conditions ? y(0) = -0.436 X ft. and y'(0) = ft/sec. 3. Now assume that oil is added to each the four shock absorbers so that, together, they produce an effective damping force of -6.93 lb-sec/ft times the vertical velocity of the car body. Find the displacement y(t) from equilibrium if y(0)=0 ft and y(0)=-10 ft/sec. y(t) =arrow_forward1. Disk Method a. y = x2,x = 0, x = 2, y = 0, about the x аxisarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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