Im 2m 3m Im 3m Question 4 10 kN 20 kN UDL = 20 kN/mD E A B 2.5m 2m 2m 3m RBY ReY 16
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could you please produce a shear force diagram for question 4
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- This is a challenging one! 10 kN 9kN 10 kN-m 1.5 m 1.5 m 1.5 m 15 m 15 m ' 1.5 m Compute the support reactions at A, C and E? [MA= 21.5kNm, A, = 2.67kN, C, = 16.67KN, E, = 5kN] %3D %3D Hint: Isolate DE, then move to BD and so on. lypforce, the sign convention is same as that used in lectures and seminars). A simply supported beam is loaded as shown in the Figure. The corresponding SFD and BMD would be: 20 kN 10 kN/m 2 m 2 m 30 kN (a) 40 kN m 30 kN 30 kN 10 kN (b) 40 kN.m 10 kN 30 kN 30 kN Click Save and Submit to save and submit. Click Save All Answers to save all answvers.solve, show all steps and fbd. do not provide copied answer or I will report Do problem 12 -22 using double integration. Also find how much additional deflection is added (to the max) if we account for the beam self-weight. Find this additional deflection using the table in appendix C. mechanics of materials chapter 12 (integration of moment)= theta, and integration of theta = v
- FIGURE PS-55 25 K 15 K 6 ft 4 ft 3.0K/ft 4 Use the free body diagram pproach shown in Sections 5-3 through 5-5 to determin the internal shearing force and bending moment al any speecified point in a beam. nd bending moneat diagrams using the guidelines pre- sented in Section 5-10.W LAB B LBC assume w = 4 kips/ft P= 5kips LAB=16 ft C LBC=8 ft What is the vertical reaction at A ? answer in kipsDraw (FBD) and calculate the support reactions in the simply support beam ABCD below 3kN 5kN 60° 0.3m 0.4m B * 0.5m 1.2m C *
- OK EFyz0 Example The figure shows two rollurs and in equilibrium. Find the valus of the reachion foras Ra and Ry of the rollers on the beam. The beam is weightluss. laterally loaded beam which is supported by a 400N 800N 200N 10ON 800 min So 000 m Solution 200 mm 300 mm 800N EFy z0 400N 200N joON 0-3 2.7 Ra tR - 200 - 400+ 100 -800 =0 Ra Rat Ry s 1300 Ra 1300- Rb ZMa z0 -(200x 0.6) + (400 0-2) - (100* 0.7) -(R 1) + (800 * ) = 0 - 120 + 80 -70 - R + 1360 -0 RE. 1360 t80 -120-10 1250 N * Ra : 13o0 - 1250 = 50 N. HW Repeat the Salution by taking the moment a boud O O General two dimen sional_fora system Fx=0, ZFyA simply supported Aluminium beam of length 18m is subjected to concentrated loads of 15 KN amd12KN ,in addition to a uniform distributed load of 50 kN/m as shown in figure 1. Assuming the weight of the beam is neglected, calculate the following: A) Replace the distributed load with an equivalent concentrated load.B) Sketch a suitable free body diagram of the beam showing all forces and reaction forces at the supports.C) Determine the reaction forces at the fixed pin support A and roller support at point B by applying the equations of equilibrium (Clearly annotate all calculations).D) If the beam in Figure.1 was applied to temperature variation Δ? such that the temperature changes from ?1=280.20? to ?2=293.40 k , find the new length of the beam (Given the coefficient of linear expansion of Aluminium is α = 23×10−6K−1).E) Then suppose the same beam temperature changes from 40 Cº to 10 Cº, explain how this will affect the length of the beam?In order to determine the magnitude of the internal forces, a FBD must be cut. It is difficult to determine where a cut should be made in order to reveal the Maximum Internal Forces, thus we have put the distance to the cut in terms of 'x', and plot the equation. In my lecture we took a generic portion of a Beam and applied equilibrium and saw that Loading, Shear and Moment are related to each through differential equations. Change in Shear is the Area under the Loading diagram. Change in Moment is the Area under the Shear Diagram. Using Graphic Integration draw the shear and moment diagrams for the given beams and loading.Problem B-3 [Design Problem – Simple Loading/2D] A simply supported structure is composed of two members AB and BE (connected by a pin joint at B). Configuration of the structure, pin supports (A and D) and external loadings exerted on the structure are shown in the figure. 500 lb/ft 30° OE BỊ Given the allowable normal stress of AB is 10 ksi, 10 kips determine the minimum cross-sectional area of AB needed 1 ft 1 ft 1 ft to carry the external loadings. Upload Choose a FileSOLVE STEP BY STEP IN DIGITAL FORMAT DONT USE CHATGPT For the following structures, calculate -Reactions -Deformation diagrams (individual and general) (6) 7 100k/b ↓ +3ft + loft 80klb T +3ft +Calculate the reaction forces on the bearing near the pulley and bearing at the far end of the shaft. Include a free body diagram and coordinates. Given the shaft length is 1 meter= Mass=40Kg Static load = 8600N Dynamic (running load) = 8400N Some hints:1. This solution will require you to draw known forces and find unknown reactions in 2perpendicular planes and then add the horizontal and vertical reaction forces asvectors to find the reaction force (hypotenuse).2. Bearings are usually mounted as near as possible to the ends of the shaft withoutfouling the other machine elements. The bearing at the far shaft end can be mountedat the 1 meter point. The bearing at the pulley end should allow clearance to assemblethe pulley.3. Bearing reaction forces are typically simplified as a single point load through thebearing centre.SEE MORE QUESTIONS