A A 200 N-m B 1 m (1) -1 m- -1 m- B 200 N-m (2) -1 m- C C
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First draw a free body diagram then determine the reactions and forces in pin B for cases 1 and 2.
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- Drive an expression for the total extension of the bar of circular cross-section as shown when it's Subjected to an axial tensile lead (P).Evaluate the following limit :(台) Lim x-2 X-2 ON4X5 5a 5b 5c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is W 8 x 13 however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? Use: Fy=248 MPa: E=200,000 MPa Designation Ag (mm2) Weight kg/m 1910 19 30 width 24 flange. thickness d (depth) mm 200.41 bf 100.06 W8 x 10 4.32 15 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narat've form to support your cho'ce) W8 x 13 2477 202.05 101.60 6.48 W6 x 20 3787 157.48 9.27 152.91 102.36 W6 x 16 3058 159.51 10.29 tf Web 5.21 thickness tw Elastic Properties mm 4 Ix x 106 mm 3 Sx x 103 5.84 6.60 6.60 13 16 17 13 126 162 220 167 mm rx 81.79 81.53 67.56 66.04 mm 4 lyx 106 1 1 6 2 mm 3 Sy x 106 17 22 72 36 mm ry 21.36 21.41 38.10244 24.54 Plastic properties mm 3 Zx x 103 145 187 244 192 mm 3 Zyx 103 27 35 110 56…
- subject: Steel StructureWhy the number of hinges=17 in this case? A XI B-C O-P XII VII D-E VIII IX X М-N VI III F 1 IV V L 2 I G H-J к п W = 3m – (3g+2h+b)=3×12-(3×0+2×17+1×2)=0 m = 12 – No. of rigid bodies; ~I - g = 0- No. of rigid joints; ~a h=17 – No. of hinges; ~A b=2- No. of links. ~1130% v - + 7 Annotate T| Edit Trial expired Unlock Full Version ENGR 263 A A A 4.8 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is a single overhanging beam. Beam AB supports a 500-lb/ft uni- formly distributed floor load on the beam overhang only. Overneia 4 Fr Stee ream (reactions) R. Free-body diagram (negligible weight) Loading condition
- How to solve this question step by step2 - Reflect & comment on how would you solve the problem if the system is subjected to ground motion? What if the spring is a yielding (bilinear or nonlinear) spring - as an example AASHTO's idealized bilinear force-displacement relationship for isolators is shown below. Force Fmax Kef Displacement Umax Uy EDCA bar AE is in equilibrium under the action of the frve forces shown. Determine T (kN). F1 88 kN F2- 34 kN "type the numerical value in two (2) decimal places only without the unit (ADD-"FOR DOWNWARD/LEFT DIRECTION, OTHERWISE, TYPE ANSWER AS IS.) 5m F1 T Sm 5m F2 5m P RA
- Compute the maximum deflection for the beam loaded as shown below 400 lb 400 lb 160 Ib/ft 100 Ib/ft 10 ft 5 ft 5 ft 3 ft- 6 ft 3 ft R1 R2 R3 R4 --- -.Use A36 steel and compute the nominal strength of the column shown. The member ends are fixed in all directions (x, y, and z). use recommended value of k. A 6.08 tw 0.282 b/t 5.87 ly 3.86 ro 4.93 no 11.5 C12X20.7 d 12.0 tf 0.501 h/t w 36.3 J 0.369 H 0.899 bf 2.94 K des 1.13 1x 129 Cw 112 rts 0.983 10' C12 x 20.78. For the plate tension member as shown in Figure 6.27 carrying axial service loads of 60 kips live load and 12 kips dead load, select the reqired thickness of the plates (A572 Grade 50 steel) the proper electrode material. 60k LL 121 DL 60% LL 12k DL 12171