5 kN/m- 4 kN/m A В 2 m 3 m Figure P-414
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- 4.5.12 OIC Section Properties Write functions for A, I, c, radius of gyration. Each function works for the three sections below. Example Function name: OICSEC_A() • OICSec_() Etc. 을 + 12 BIP - bhs V, 12(BII-bh) BII -- bha 6/1 - 8An element in pure shear is subjected to stresses Txy =-14 MPa , as shown in the figure. 14 MPa Using Mohr's circle, determine the principal stresses. MPa (f) ơ1 =| (g) 02 = MPа (h) Show all results on sketches of properly oriented elements. Choose the correct sketch. P 14 MPа 14 MPa А. P O, = 45* 7 MPa 7 MPa В. P = 45V M R = R R = D RA M 40 kN = VD₁² M VD2= MB1 B2 D = 2 m 6 m MB2 MB1 hinge 4m-4m VB Vpz KN KN 2 kN/m KN KN KN KN kNm kNm kNm D RD VD1 MD 8 m 2° 30 kN-m RE
- H.W: draw the I.L for RA, RC, RF, RD, MD, VCL, VG and MG. A 2 m B -2 m C G 0.4 m 1.6 m D 2 m- E 2 m LLUšing briigratian, detimire ka canchord of the X-4130% 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
- Тв d Tp (1) (2) TE (3) (4) 1A C D id L2 L3 L4 re FIGURE P6.14 gs ge A.B2-8m RC Beam Sec B-B Sec A-A Sec A-A C1 (0,8x0,8)m Sec A-A- Sec 8-B A Sec A-A- 143.9* Sec A-A Sec A- Sec B-B Sec C-C Sec C-C Soc B-B Sec Sec C-C -3.185m Sac A-A-D 83=4.5m Sec B-B -Sec 8-8 C1 (0,8x0,8)m DBT Figure 2: Plan of Reinforced Concrete Beams Sec A-A -C1 (0.8x0.8)mS ( x³ + 4x² ) dx 1. (e –1) dx (e +1) 2.
- A 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 RADesign a signly r/f section for BM 100 KNm M30 FE 500 EC 50 MMIntegrate the following: ,S (xª – 5x³ + 6) dx - 1. 6. su(1 – dx 2.