For the frame shown, E = 30x106 psi, A = 8 in², and I = 800 in for all elements. Determine (1) the nodal displacements and rotation at node 4; (2) Reaction forces in each element in GLOBAL coordinate system. (Do NOT break element 3 into two sub elements) T 20 kip 25 ft 25 ft 40 ft 20 ft 20 ft 30 ft 3
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- h B h + d + d d THE FIGURE IS NOT DRAWN TO SCALE. Also, A and E are BOTH pinned (hinged) supports (even though they're not drawn that way). Numerical values for the parameters shown are P = 680 N, d = 0.40 m and h = 0.40 m. For component directions, use x positive to the right and y positive upward.In the graph the force versus displacement of a spring is given (the spring is shown in a separate figure--see below). The x-axis range is ±2 cm. The y-axis range is tFs, where Fs = 220 N. How much work does the spring do on the block (the mass "m") when the block moves from x₁ = 7.6 cm to 4.3 cm? W J y F. X -1.5 -0.5 0 10.5 1 1.5 -F How much work does the spring do on the block (the mass "m") when the block moves from x; = 7.6 cm to -7.6 cm? W = Question Help: Read Submit Question m3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)
- Problem 4.61 Figure 15m 0.5m B 2.5 m 1 of 1 Part A Determine the magnitude of the moment of the force F = {50i- 21 j-80k} N about the base line AB of the tripod. (Figure 1) Express your answer to three significant figures and include the appropriate units. MAB= Submit μÀ Value Request Answer < Return to Assignment Units Provide Feedback ?:Q81 The moment M for the force the figure shown below is 250 N 153 -200 mm 30 mm 46.5 N.M O 46.6 N.M O 46.4N. M of L isOf a column of infinite length to points A, B, C and D parallel FA, FB, FC and FD (kN) respectively Using the forces in the directions given in the figure, equivalent resultant force (FEŞ) size and application x and y (m) coordinates of the point find. FA=20 FB=55 FC=35 FD=77 xA=13 yA=8 xB=3 yB=5 xC=10 yC=0 xD=-9 yD=-15
- 8:30 Test 3.pdf F1 F2 a CF3 A b a F4 D Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30°, 0= 45° , B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m e) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m 0 None of them 21 What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 KN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m ) None of them 31 What is the resultant moment of the five forces acting on the rod about point C? b) - 21.1 kN.m e) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m ) None of them a) 32.6 kN.m 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m ) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e) 100.7 kN.m ) None…Review Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. Part A - Moment due to a force specified by magnitude and endpoints The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo =r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are ¤1 = 0.350 m, x2 = 1.90 m, y1 = 2.30 m, and z1 = 1.20 m. What is the moment about the origin due to the applied force F? i j M =rx F =|rz k Express the individual components of the Cartesian vector to three significant figures, separated by commas. ry F F,…1. - solve the nodal displacements and reaction forces: Al 20 kN/m 20 kN/m 20 kN/m 20 kN/m 5 kN ww Nom
- Choose the right Free Body Diagram representation for the Ring shown Select one: A H = 0.15 Hk = 0.12 20° 6" В Hy = 0.15 ek =0.12 20° C 6" H = 0.15 Wk = 0.12 20° 6" Hs = 0.15 Hk= 0.12 20°Q1: In the below drawings, three unbalanced masses and related details are provided. Please answer the following questions. a) a balancing mass is to be located at rp = 50 mm, then what would be the balancing m1 m1 650 mm 200 mm m2 mass (m,) and its angular position (Op) (Please use the r2 m2 A graphical method) r3 m3 15 m3 b) If the system is not balanced what would be the reaction forces at A and B. (FA, FB=?) n = 150 rpm r; = 30 mm m, = 1 kg r2 = 20 mm 40 тm 13= m, =3 kg m; = 2 kg k = please choose an appropriate scaling factor yourself SHOT ON MI 6 MI DUAL CAMERA9:10 IM N 10 36 | zain IQ 22222.jpg > SAMPLE PROBLEM 2/5 Calculate the magnitude of the moment about the base point O of the 600-N force in five different ways. 600 N Solution. 2 m 600 N F- 600 cos 40 4m F= 600 sin 40