QUESTION 1 Figure Q1 shows a frame structure that is connected to a cable with a tension force of 150 N. Determine the resultant internal loadings acting on the cross section at points E and F of the frame. 3 m 1 m I m F 100 N/m B E -2 m- -2 m- 1 m 1 m 30° Figure Q1 150 N
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- Q13. As shown in the image below, the member is subjected to the applied loads: a free couple moment M=320lb*ft, concentrated force F1=380lb, and concentrated force F2=220lb. The dimension a=5.5ft. Determine the magnitude of the resulatant moment (in lb*ft) caused by these applied loads about point A. Answer must inculde 1 place after the decimal point.Please:the answers are fbc=1400N ax=1200N, ay=-800N, Moments(this was forgotten last time) about a Max=-600N ,May=-1200N, Maz=-2400N
- A weight W = 905 lb is held up by two cables each with a diameter of 1.5 in. Determine the tension in each cable and the normal stress in each cable. 0 = 21° O = 37° B Image is not drawn to scale.The timber member has a cross-sectional area of 1703 mm2 and its modulus of elasticity is 13.2 GPa. Compute the change in the total length (in mm) of the member after the loads shown are applied if x = 1.84, y = 3.63, and z = 4.11. Round off the final answer in two decimal places.Mechanics Of Materials I
- If the tension in cable DE is 455 N, determine the moments of this tensile force (as it acts at point D) about point O and about line OF. EQ 2.3 m Answers: BC=CD = 2.2 m MOF= 3.0 m Moi -1203.39 i 2.3 m 3.1 m 1.3 m i+ i N.m 1.7 m 39° j+ i k) N-mForces shown on the diagram please answer all partsQI/ The rigid pole and cross-arm assembly is supported by the three cables shown. A turnbuckle at D is tightened until it induces a tension T in CD of 1.2 kN. Express T as a vector. Does it make any difference in the result which coordinate system is used? Ans. T = 0.321i + 0.641j – 0.962k kN, 1.5 m /C B 1m 1.5 m T = 1.2 kN 3m E 01.5 m 3 m 2 m m