For the beam shown, use only singularity functions. V₁ = 50 lbf/in and V₂ = 14 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. O R₁ 4 in 400 lbf Hinge JA B AC R₂₂ 4 in 2 in V₁ | V2 D R₂

For the beam shown, use only singularity functions. V₁ = 50 lbf/in and V₂ = 14 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. O R₁ 4 in 400 lbf Hinge JA B AC R₂₂ 4 in 2 in V₁ | V2 D R₂

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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My Courses This course EHide blocks on C- Fall 2020-2021 > General > MENG250 - Midterm Exam - Fall 2020-2021 Question 4 The beam shown in the figure is subjected to two forces. What would be the resultant moment about point A produced by the two forces. Not yet answered F = 375 N F, = 500N P Flag question 8 m 6 m 5 m- Select one O a. 875 N.m O b.4589 N.m O C-8600 N.m Od. 2600 N.m A d) G D ENG 7:54 PM 12/11/2020
The answer must be wn= sqrt(3g/5b). Solve this without the application of Moments please.
Use the diagram shown above. Given: • D₁ = 10 ft . • D₂ = 3 ft A #/ft . W = 100 W-100 Solve for the internal moment at X. D₂ D₁ W X Note: Enter your calculation results without the unit. You'll be asked to specify the unit for your calculation result in the following question. (e.g. if it's positive 100, enter 100, if it's negative 100, enter -100.)
Brandon is trying to create a cool shelf bracket using pipes like he saw on Instagram. He is concerned about the pipe unscrewing from the wall bracket at A. He wants to know how likely it is that it will unscrew. Using some combination of drawings and written statements, answer the following questions. 1) What moment direction/axis would be critical to the pipe unscrewing at A? 2) Based on the current geometry, which force direction of F is most likely to create the moment described in part 1? 3) If the weight of the books acts only downwards, how could he modify his bracket so it cannot unscrew? You can use any combination of the parts shown (no need to use all of them). Show where the now vertical load F will be applied. Explain why your new design will be less likely to get unscrewed at A. x B 0.5 m 0.4 m A 0.3 m 0.3 m
In your response, please display the full computation and all FBDs that were used. Only round off at the final answer with 4 decimal places. P=2kN, F=3kN, and moment M=0.5kN-m are acting on the structure in the illustration. A pin at Z and a cable RQ connected by a connecting pin R support it. Im ● F ● P S TO M Im R lu W OZ a' 0.8m 3m If the pin support Z's factor of safety against shear stress is 3, determine whether a single or double shear should be provided for an economical and safe design. 4 m The mas allowable shear stress in the pin is 50MPa. Consider the following specifications. Single shear design: 25mm pin diameter Double shear design: 17mm pin diameter
Question 13 Use the diagram shown above. Given • D₁ = 9 ft D₂=7 ft D₂ 3 D3-4 ft • W = 100 #/ft P = 800 # D. 1 D₂ X W P Solve for the internal moment at X. Note: Enter the numerical value only for your calculation results. You'll be asked to specify the unit for your calculation result in the following question. (This is due to Canvas' lack of ability to require units for calculation question.)
In (Figure 1), w = 3.8 kip/ft and F=20 lb. Follow the sign convention. Part A Determine the shear force acting at point D of the beam. Express your answer in kilopounds to three significant figures. vec VD = kip Figure 1 of 1 Submit Request Answer Part B Determine the moment acting at point D of the beam. Express your answer in kilopound-feet to three significant figures. A, 6 ft 9 ft ΑΣφ vec MD = kip ft %3D
3) 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.)
Question 1.9 Qu A beam is constructed of a channel section. The cross section has the dimensions indicated in Figure Q19 and the moment of inertia (second moment of area) about the z axis (the neutral axis) is I=1.2 x 10 mm. The bending moment at a certain point is M = -1 kNm. The maximum tensile stress o,and maximum compressive stress o resulting from this bending moment are 15 mm 2- H C Iss mm Figure Q1.9 ot = 29.2 MPa and oc = -29.2 MPa ot-29.2 MPa and oc = 29.2 MPa O C. ot = 45.8 MPa and oc = -12.5 MPa O d. ot 12.5 MPa and oc = -45.8 MPa O a. O b.
Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Let a=4.0 ft, b-9.0 ft. c=4.0 ft, d=3.0ft, w=8 kips/ft and P = 66 kips. Construct the shear-force and bending-moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. B B ÎÎÎÎÎÎ Answers: Ay=i b d For this loading, calculate the reaction forces Ay and Ey acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free-body diagram below. (Note: Since Ax = 0, it has been omitted from the free-body diagram.) W C b C C D E kips, Ey = i E X Ey kips.
Im doing a shear and moment calculation for a problem graphically and keep getting a negative sign in my moment calculation where I should be getting a postive sign. I know that the integral of my shear calculation is going to be my moment equation but I always use this to "check" myself against my work after I do my standard summation of moments about a point. I KNOW I SHOULD be getting M_2(x) = -.5x^2 + 2x + c but as you can see by my work I keep getting a "-2x". Why is this and where am I going wrong?
Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Let a = 2.8 m, b = 5.6 m, c = 3.8 m, P = 20 kN, w = 45 kN/m, and Q-10 kN. A P B W b C X What is the maximum bending moment on the beam? Answer in kN-m rounded-off to 2 decimal places When entering your answers, use the bending moment sign convention. I