013P 6-19. Draw the shear and moment diagrams for the beam. 2 kip/ft 30 kip-ft 3.F A 5 ft 5 ft 5 ft

How can we obtain the maximum bending stress using the flexure formula?

Part (b)

Activity 3 A rig is to be constructed, which will consist of a 10m long elastic beam, simply supported at both ends, as shown in Figure 1. The first simply supported beam is subjected to the concentrated loads and the second beam is subjected to the distribution and point loads. Determine the distribution of shear force. Calculate the bending moments, due to bending in the simply supported beam. 500N 1000N 2000N 500N 2m Im lm lm Im 4m Figure 3: SSB with point loads 80 UDL: 10 (KNm) 2 (m) 7 (m) Figure 4: SSB with the combination of point and distribution load

For the simply supported beam and its cross section is shown, Determine: 1) The S.F. and B.M. diagrams; include the mathematical solution steps of S.F and B.M. 2) The position and magnitude of the maximum shearing force and bending moment. 3) The location of any point of contraflexure. 4) Determine the value of the side of the square ( a ) if the safe bending stress in the materials is equal to 100 MN/m. 5) Determine the maximum shearing stress in a section 1.3 m from A.

FIGURE PS-39 25k/h 2t 4ft niques shown in Section 5-2 2. Draw the complete shearing force and bending moment diagrams using the techniques shown in Seetions 5-3 through 5-8 3. Determine the magnitude and location of the maximum absolute value of the shearing force and bending moment. 4. Use the free-body dingram approach shown in Seetions 5-3 through 5-5 to determine the internal shearing force and bending mement at any specified point in a beam. 5. Write equations for all segments of the shearing force and bending moment diagrams using the guidelines pre- sented in Section 5-10.

Figure 1 Q2- Draw the influence lines for the vertical reactions at supports B and D, shear and bending moment at point C for the given continues beam shown in Figure 2. determine the maximum positive and negative shears and the maximum positive and negative bending moments at point C due to a concentrated live load of X₂ kN, a uniformly distributed live load of 2X₁ kN/m, and a uniformly distributed dead load of X₁ kN/m. A X1 = 100.8 2X1= 201.6 X2= 549 -5 m- B -6 m- C Figure 2 -4 m- D d -3 m- E -4 m- F

6. Finds the bending stress and equivalent bending load due to bending of a 6 x 19 (Fiber core)  wire rope of ½ in. diameter around a 24 in. pitch diameter sheave.

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Q3: For the load diagram shown in figure (2) below, the first contraflexure point of the bending moment diagram is located at ... m from A. 50 kN 60 kN/m 8 KN.m 2 m 4 m 2 m Figure (2)