psad sample problems, Exercises of Engineering

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CERTC Rm 501, G.K, BASIC STRUCTURAL ENGINEERING Lecture Exam 1 INSTRUCTION: the following questions. each item by shading the STRICTLY NO ERASURES ALLOWED. MULTIPLE CHOICE SITUATION. REFER TO FIGURE 1. A circular hole is Select the correct answer for each of Mark only one answer for box corresponding to the letter of your choice on the answer sheet provided. PSAD-BASIC STRUC COMPLETE ENGINEERING REVIEW AND TRAINING CENTER Chua Bldg., Contact No.0917-302-8824 M.J. Cuenco Ave, Cebu City 10. Calculate the modulus of-elasticity (GPa). R. 247 B. 175 182 D. 154 11. Calculate the ultimate stress (MPa) . A, 978.7 B. 294.7 “0 486.3 D. 247.6 SITUATION. (APRIL 2025 CELE). REFER TO FIGURE 3. A system of pulleys is shown in the figure. to be punched in a plate that has a shear 12. If W2 = 500 N, determine W1 (N) for strength of 40 ksi. The working compressive equilibrium. stress for the punch is 50 ksi. 8) 325 c. 150 B) 125 DB. 250 ee oa bien of the plate in 13. If W2 = 750 N, determine the tension (N) in a a ameter can be punched. the cable between pulleys A and B. . O. in. c. 0.80 in. 6) 0.78 in. D. 0.82 in ae ©) are 7 Oe : HB. 450 250 If the plate is 0.25 in. thick, determine the 14, If W2 = 750 N, determine the tension (N) in diameter of the smallest hole that can be punched. ie cable at the right of pulley C. (a) 0.80 in. « C. 1,20 in. ®) 187.5 Cc. 375 B. 0.60 in. D. 1.60 in. B. 450 D. 250 3. A spherical shell with 70-in. outer diameter SITUATION. (NOVEMBER 2023 CELE). REFER TO FIGURE and 67-in. inner diameter contains helium at a “4, Shown in the figure is 2 6-mm thick Pressure of 1200 psi. Compute the stress in the cylindrical tank with a diameter of 1.2 m. Tt is shell. subjected to an internal pressure p = 0.9 MPa and A. 26,800 psi B. 28,000 psi 1) 3,400 psi 14,000 psi SITUATION. REFER TO FIGURE 2. rectangular wood panel is formed by gluing togethe two boards along the 30” seam. 4, Determine the normal stress in the glued joint— P= 10 kN. 1 ¢ orm kea C. 28.6 kPa +o | 49.5 kPa D. 24.7 kPar \ 5. Determine the shear stress in the glued joint The 250 mm x 700 mn er Pep is supported by 16 bolts as shown. q.® « (MPa 457 16. Determine the tangential or circumferential paw Stress (MPa). AL 70 As) 90 17. Determine the allowable internal pressure p (MPa). Bolt tensile capacity = 52.5 kN each Factor of safety = 1 15. Determine the longitudinal stress A. 70 B. 90 c. 35 35 D. 45 J if P = 10 kN. P A, 1.41 B. 0.37 4c) 0.50 D. 0.74 A. 42.9 kPa C. 28.6 kPav The Ot B. 49.5 kPa (2) 24.7 kea Not tgiqva-s SITUATION. (PAST BOARD). A solid steel post that Ne SITUATION. A i5-mn-diameter steel rod which has length, L+15 nm, is subjected Lo its own weight. An ore bucket with weight P is attached at the end of the rod. Unit weight of steel = 77kKN/m3. 6. Calculate the value of P (kN) based on 12-mm is free [rom one end and fixed from the other end 4 Anasguyis Stbjected to pure tension ait Given: Post diameter = 80 mm Length = 3 m Shear Modulus = 70 GPa 18. What is the torsional rigidity of the I-mi?) 2 post eee c. 28.27 L ao) 281.49 kN-m? Cc. 235.15 kN-m? ) 38.07 D. 30.68 es 266.91 kN-m? D. 224.67 kN-m? If p=20 kN, find the longitudinal strain. ki: fL — the torsional stiffness of the “Wy 0.00087 cae 4 AE AL 99.62 kN-m —©) 93.83 ku-m acd Os B. 69.95 kN-m D. 96.97 kN-m at] 8. Find the ductility of the rod if due to a 5 20. Det: the maximum shear stress (MPa) in heavy load at failure the diameter of the broken fein post it ir oe subjected toa Hennes S40 Ne rt is 14.50 mm. a) 6.568 0.3.458 5 3 oa Does A) §.37 MPa C. 6.15 MPa 9 . Th. oe . 6.25 MPa D. 5.61 MPa Je SITUATION. Use the following data to answer the Giwarion. REFER fo Ficuee ©. por the Gezesa following questions: nape thee . Initial diameter = 12 nm Length = 50 mm element shown in the figure. Diameter at rupture = 11 mm 21. Determine the major principal stress in kPa. Load at proportional Limit = 28 kN A. 174.4 c. 144.7 Elongation at proportional limit = 0.068 mm B. 147.4 D. 114.7 Load at rupture = 55 kN Length at rupture = 58 mm 22. Determine the shear stress in kPa on plane cD. 9. What is the. luctility (%} of the material? AB. 69.3 c. 55.3 A, 8.33 ‘B) 15.97 c. 19.01 D. 9.10 B. 44.7 D. 32.5 p Figure 2 . Figure 1 Punch A Metal sheet B q P 120 Pa We Figure 5 redo bPa c Wz — c0 kPa Figure 3 § Figure 4 f & Norn force, By = Peos70" L, ests’ U = Shear force, Py= Yéing0" BNR ? 4 x R= Paste" 7) - (980)(x) Fe »_ 700. pie. A } < (oso _W, ee 4 = (200) 399, ia 7 i In 2 410540 = YOu Paz We - Leos a Test) _ 300 Uy y . ed 19) Xs 2a" i? = 5 29 Wl ee ee [ i P6008) t ie (wooo) al pe Lak 4 Sw FSH) AEE) w= 77 , 000 rc ( v8)(55) SW avd 9® ; rn ) €- et a aa 0 4 iar “eon 1 wo) j ( Ke $.53-- -® “4 AS-s 5,2 x10 & 6,000 a\o.owSt} (ooo 8054 +S Yookels Law : E= € 20 00042) 2,000 = ~~ { £ = 0.00087 jer oe | ul ; pete ae Ace oe tly ¥)}° fm | _ gia ang rapture fs 7 J ifr v xR = 0), Q mu " / G7 00 | a = rn Past” Cc IN | Lephn cS nm Ai) wah f elas, = 2 ! Straw) ; one A800 styyss= Sk Te >A) Mea area qa Si == 1 2 I36KI0” ft 192, 039.9% May = |I92 GPa 6) t-Cnm d= 12m = 4,20 man Par 09 WE aC) bolts OD Fe - OIG? we oY 4 =a LYE > BE . 0.5{leo9) _ D seduction sactor : b Tn “Ry S —" MD) 74 16) - \90 ry) on “ 7 ( () facer of Saget Py -2y me (0) D pet aan = Sk kN ASS 2 R. Sto oss tal Fo, (rt) Fog (retaeyte Sau? W ,+ 244% bap ~ Lal) & poor momar! of tertia: J > Txtly - 10) orsimal stypness =~ © ree ret OO 4\ lagc a [Om bso) a Tee a tusinnal Shear stem T bo shor diye = 30 GM 2M get) = 400 x10 ont ats! a ty = JG pa EL x ongfe 9F Twist, 4 Je + dasa fjditys J6= © ag (@00@ 32) d | a? a toRdonal Tigh d v + Llexura! 9, = 261g x Oe a axial gigiirty = AE 4 kN wy)