(a) The figure below (figure 8) shows a non concurrent system of forces acting on a beam. Consider the equilibrium conditions and calculate the reactions at support. Take the values of P and Q as 8 KN and 4KN respectively. y P 15 kN·m 10.5 m 1.5 m 1.5 m-1.5 m-1.5 m figure 8
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- Solve the preceding problem for sx= 11 MPa and ??y= -20 MPa (see figure).The fixed-end bar ABCD consists of three prismatic segments, as shown in the figure. The end segments have a cross-sectional area A1= 840 mm2and length Lt= 200 mm. The middle segment has a cross-sectional area A2= 1260 mm2 and length L2= 250 mm. Loads PBand Pcare equal to 25.5 kN and 17.0 kN, respectively. (a) Determine the reactions RAand RDat the fixed supports. (b) Determine the compressive axial force FBCin the middle segment of the bar.The bar ABC shown in the figure is loaded by a force P acting at end C and by a force Q acting at the midpoint B. The bar has a constant axial rigidity EA. (a) Determine the strain energy U1of the bar when the Force P acts alone (Q = 0). (b) Determine the strain energy U2when the force Q acts alone (P = 0). (c) Determine the strain energy U3when the Forces P and Q act simultaneously upon the bar
- Problem 3. As shown in the Figure below, please determine the internal force in the middle of AB, with the given data: OA-15 cm, OB-30 cm, OC-60 cm. W=40N, Wo=60N. F=1214N, B-10°, FM=1236 N, and 8=15%. Fj B FM W Wo 4A rigid beam is supported by a pin at A and two metallic wires at B and C. Determine the force P that causes the point C to displace downward by 0.1 mm. Given: E (wire B) = 200 Gpa, E (wire C)70 Gpa and both wires have a diameterD =6 mm. Consider a linear elastic behavior.A rigid beam is supported by a pin at A and two metallic wires at B and C. Determine the force P that causes the point C to displace downward by 0.1 mm. Given: E (wire B) = 200 Gpa, E (wire C)70 Gpa and both wires have a diameterD =6 mm. Consider a linear elastic behavior.Determine the value of reactions at the supports A & B. see Figure2. Figure 2 15 KN/m 1ORN/m Brn 4m 3m
- A distributed force and a concentrated force act on the 20m long bar shown below. 2 kN 400 N/m A B * 5m * 5m - 10m Draw a free-body diagram replacing the distributed force by a concentrated force. Calculate the three support reactions. (Result check: Ay = 3.5kN)The bars AB, CD and EF made of material with E=122 GPa and have a cross sectional area of 617 mm2. The beam ACE is subjected to the force P=86 N. Take L=1.4 m and d3596 mm. Answer the following questions D E 1/The force in the bar (AB)is ? 2) The forcein the bar(cD)is 2 3) The Porce in the bar CEF) is? 5 The stress inthe bar (A BJIS? 5) The stress int he bar cD)IS? 6) The stress inthe barcEFJ is? 2) The elongation inthe bar(AB)IS? 8) The e longation inthe bar Ef)iS?The simply supported beam is subjected to the loads as described in the figure and parameter table. Neglecting the thickness of the beam, calculate the horizontal and vertical support reactions at A and B. F A B C -4*- L3 parameter value units L1 L2 L3 ft 4 ft 6 ft theta 65 F 200 lb M 500 lb-ft For the answers, take to the right and up to be the positive directions. The support reaction at A in the z direction is NAz- lb The support reaction at A in the y direction is NAy- lb The support reaction at Bin the z direction is NB= lb The support reaction at Bin the y direction is NBy= lb
- An aluminum rod is hanging from one end. The rod is 1 m long and has a square cross section 20 mm by 20 mm. Find the extension of the rod resulting from its own weight. Take E = 70 GPa and the unit weight = 27 kN/m3.Axial loads are applied to the solid cylindrical rods as shown. The loads P = 1500 lb, Q = 900, and R = 1300 lb. Find the maximum internal normal force in the structure. (A) - 2300 lb R (B) - 1500 lb (1) (2) (3) (C) - 1300 lb R (D) - 300 lb (E) O lb (F) 300 lb (G) 1300 lb (H) 1500 lb (1) 2300 lb (J) None of the aboveM y A 2L - 30% В Calculate the support reaction at the roller at A to the nearest whole number. Use up and to the right as positive, down and to the left if negative. M = 63 KN-m P = 143 KN x = 2 y = 7 L = 5 m