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ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 STRENGTH OF MATERIALS AXIAL STRESS AX|AL STRAIN RELATIVE ae MENT _ f + Sag = SB 7 Sa 6 A ¢ L es \ toL BEARING STRESS HOOKE ‘Ss LAW Sign Convention: P y oL * Sie deecined 6p = = = ve Ing Named beg + 4, Re SHEAR STRESS AX|AL DEFORMATION PO|SSON’s RATIO Single Shear ¢— PL _ oL y = clatral € J: NU AE E longitudinal & Ay Note: | yan 2 Gar | MODULUS OF RIGIDITY Double Snear if due to = GY -y — shear drain 5 | Solt weignt G-E- Relahonmip 2Ay Relation.nips : ; E Punching Shear OX € 2(1 +) q.—V O= ke BULK MODULUS OF ELASTICITY Ramer? t k= E E oO om ae 7 = E- € 3I-2v) Vi Qe: ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 STRENGTH OF MATERIALS TORSIONAL STRESS lt] . >. tH ee POLAR MOMENT OF INERTIA J? Tx + Ty ° for Circular Sections (Solid) distance of boltjueld (fr.centrnid) T= F(R) = 5 (py © for Circular Sechons (Hollow) lee ey J 3 (D-d") TORSIONAL DEFORMATION (Angie of Twist) TL JG Angular Deformaton Pap = Be - Ba § = Maximum Torsional Shear Stress Solid : Shalt Jmox = 1D? IGT Tmin = O Hotlow: halt Jmax = r(D* d’) IbT D TORQUE DIAGRAM (0) Get reactions @ Cut @ Lett & Right of each jt. @) Analyze left side note: CW > COW G (4) Formulas re —_— Tae = Te pt. J -"f Bis = JG (from A- B) MOMENT DIAGRAM (rl ee) Point of Zero Shear Shear Xpzs = M Pzs load ( max ) Point of Contraflexure x@ M=0O (Zevo Moment) + Shear 7 - Shear Curved Beam (ean) away from center 4oward center Spandarel Propeyty n- ( n+l bh) n= degree x= — (pv) n+2 <— considering vertex ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 STRENGTH OF MATERIALS FLEXURAL SHEAR STRESS PRESSURE VESSELS Rectangular ° Thin - Walled 1. 3V F = Pret Pia L | 2h hee | | Circular ° Circumferential / Tangential / an (Solid) (Hollow) r+ Paet Din SA A * Longitudinal / Spherica| 5, = Pret Din MINIMUM Merax 4t 1 Pt. Lift e Shear Strecs X= 0.2929 L (Mmin) T x= 02071 L | (iin * Snear Flow (a a 3 Pt. Lift 2Aw | X=O.M5L (Monin) Note: Min@ |Mpos| = | Mneg | ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 ENGINEERING MECHANICS Analysis of Cables a) PARADOLIC CURVES Tp (symme Meal ) Teak = (TA F = wl (Tnox)* = (4E)'+ (4. )’ H= a = minimum tension @ lowes! Pr. for unsymmetrical; familiarize Rg P and Me process. 0) | CATENARY WAVED Length of Catenary Cables fo” We To Tensile Force at any pt. of the cable T = Ts cosh (= Wes We) To_ siah (e+) Vertical distance of the lowest to any Pr. oF the cable y= | cosh (5) | where: T,) = alway horizontal x +o any pt. OF the cable _ horizontal distance of lowest SYMMETRICAL Lv L.-T, y-s” =¢* Ne) wW=ws ke y= c cosh (%) To = WC UNSYMIMETRICA L T = wy r 8S = csioh (2) Ta = W Ua Te = WY H = wC L = Xa t Xp Sr= Sa + Se DYNAMICS scalar Speed —— distance velodty — displacement — vector CVURVILINEAR MOTION Tangenhal / Uniform * a; — trom 3 eqns Novmal / Radial /Centripetal z Vv Ou RR Toral Acceleration ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 ENGINEERING MECHANICS RECTILINEAR MONON a) Unifowm Accelerahon Ve = Vo tat Ve’ = ve + Las S= Vot + sat* b) Uniform Velouity (a=0) S=Vyl c) Variable Accelerahon | S - position . A da ve-dS a= Ms (sep rT 1 (JERK ri | Vdv = ads | > By Graph ° Using a-t diagvam Ve = Vi + area@time t 5 = wt +[area@ time t + xq | e Using 5-t diagram AS = Change in area of the v-t diagram ROTATIONAL MOTION Wy = Wj tat. Wy = Wir t Jn0- 8 = witt Fat where wr=v LINEAR- er = 3 j ANGULAR ar=a RELATION PROJECTILE MOTION @ Summit, vs =O z Ynax = H= 2 @ Some Pt. Along he Path t y = xtanp - 22 _ 2vi* cos*® NOTE: max possible horizontal vange = fired at 49° NEWTON 's SECOND LAW OF MOTION F=ma equilibrium f dynamic / in motion a#0 , Minimum gpeed for objects pasving thru | - [oR Note: coneiant path (T= 0) ee V=¥gR WORK- ENERGY (KE + PE Jinn + Work = (KE + PE), where: PE,= mgh = Wh + Gravitational PE PEe= ky’ Elostic PE KE = zm" + Kinetic Enevgy Work= Force * displacement ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES STRUCTURAL STABILITY © Frames : IMPACT LOADS Beams |R-3-9|| R-G-C ene IN oop -5| 5 | GNiooe-S S, = Ser ( a) Trusses |M+R-23||M+R-3T © Trusses : Gor Frames 3M +R-(37+3), OM+R-(67+S) SD mtr = 3 where: R= number of veactions | | pf vaive: I ST mtr 7 4 MOMENT- AREA MTD. 5- Tange” inte vaforts =O : SD “UnSt. =mty < 2 Theorem 1 M= number of members 70 : ST f J= number of joints <0: unST.| POUBLE- INTEGRATION MTD. Pen = Er (Area wae) Number of Releases, S @ Create the Moment Eqn. 6 an Nengent Ron 0 MTP Intemal tinge | { ‘Ely'=M); Rigrttut sow Theorem 2 [ternal Roller 2 @) ntegrate twice tbe L (Area =) c, Internal Fixed =| O Ely=SM +Ci| slope eq’n. EI a Slider 1 Ely= JIM + C.xt Ce, deflection eqn solve Seagal ot © Kdeeiaron) | Techniques @ Set Boundaries for constants € Solvable via Rat & {roportion. @) Beams : SIMPLY SUPP: (YO) ;(X=/0)D) WorE:G (CC) o° SD or = 3n | [wheres Fixed: GY =0);(G=0);x= OD 4 ad l° A= an (th) SIT or 7 3n [ir fumber of eax. Max. Deflection: y=0) $11 A 2° \ivaegb Unst. or < 3n_ |n= ter oF memPas a f 3 ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES MOMENT DIAGRAM BY PARTS Loan | DIAGRAM | AREA | CENTROID| HEIGHT. h f ee L Ga );oC= Lh t c P = ae Lh L : yf Lh L wit EPH |)/——~=) / 3 5 w = et} kh Lt wl" yy | 4 5 6 STEPS : @) Sowe for reactions (@) Apply Moment vy Farts at Fixed Support @) Draw the Elastic Curve @) Apply Moment- Area Method Theorems NOTE: UPWARD LOAD = positive (+) DOWNWARD LDAD = negatwe (-) CONJUGATE BEAM MTD. Real Beam Conjugate Beam XS A Steps - Q) Solve for reactions (@) Set location of the clamp (3) Draw moment diagram (4) Cut and isolate (6) Slope 1. Shear 4 ET (G) Def lection y= Moment ET ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES FIXED-END MOMENTS POINT -LOAD ANALOGOUS (@) Convert the areca of the load to concentratd load as vertical «trip (&) Use formula for concentrated Joad then integrate (©) The limit of iakgvaton whould be the bounday of the area P= ydx Eee ie +———_.—— TABLE FOR CANTILEVER Nok: For any loading ML 2 L ay\t% 2 ray - ff eae — i ML 4 rays frente — ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES SHORTCUT TABLE FOR DEFLECTION AT MIDSPAN FIXED-END SPAN SIMPLY SUPPORTED SPAN D mia = E 4 0.50 w O and/or ZFy=O ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES INFLUENCE LINE Muller Brestaus Principle Reachon: 2 1-0 —b clo b as ab vanite, ZS Propevties to Use: Function = Py, + Peye Function = w,:A, | influence Line for CANTILEVER Shear: | | f° Moment: S| 4 -L MOVING LOADS et P= P+ Ps & L For Maximum Moment, _ (PL-Psd)° 4PL or:(@) Get the resuHant load,P. @) Locate P via Varignons Theorem ©) Position the center of Peand P at the midspan of the beam @) Compute the moment under Pt Influence Lines of Simple Beams Re Ra | | | = Influence Line for Ry T Influence Line for Rz ‘b/t | - Length of | : | simplespan ! —— ple span | ol -a/L Influence Line for V- For Maximum Gear, Influence Line for Mc 4 R=RtPs 2 + ce) Vane = PL-Fed R = Vax one LL OR: @) Place the bigger load under a support (@) Compute the reaction under the biggest loac () Max shear = Reaction under the load ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF SIRUCTURES Influence Line for TRUSSES SAMPLE: Influence for Member GH | a=3m b=9m -(a)(b)/LH = -0.75 H=3m 1 SAMPLE: Influence for Member CD “-+(a\(b)/th = +1.0 L=12m | “P= 10 hinged NOt. Check if there is reduction factor for momevit of inertia. (2) Get the value OF k using the Nomogram Chart AS 1 Ha a ig a # = = ao sie 3 £ 50.0 co a a oo 4 = Lo a 100.0 100 4 E 100 oo | 20.0 Figo F 50.0 50-2 p60 30.0 4 > 50 F- 30.0 204 a oe - 3.0 20.0 4 4.0 — 20.0 204 B 4 + r wae L. 10.0 4 iti 19.0 + 3.0 te 19) ao 4 rT EK 20 = t— 7o4 - 7.0 4g =| oy 40 aa 60 a7 L~ OF 5.0 4 + t- 6.0 | 06 4 oe ae — 40-4 =~ 2.0 = 4.0 ne E 304 T L-3.0 o4 C 4 IL b. 03 - 4 + L- 2.0 ae J +18 L 4 r + + L 1.0 r E vail —Los Lo = +10 Le @) Solve for Slenderness Ratio ; aad op= KLu fi where! r= 0.284, ectangular r= 0.25D cirevlay PATTERN LOADINGS for TWO SPANS % Max. negative moment © mid- vupport \f Both Span is LOADED with “u)” wi* g If One Span is LOADED with “w% Mneg = Z Mueg = = ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 THEORY OF STRUCTURES Addihonal : FIXED- ENDED BEAM: For any loading FIXED- ENDED BEAM with added support at midspan @(V =0) +Mmax= R= (w) [2 24 lh R= (WG) wl? +Mingx= 4 @ (V = 0) SIMPLE BEAM with added support at midspan wl? —Minax= -“— 8 I 3 R=S()O +Mmax= + @(V = 0) w I R=" L = 5 (w)(L) owl? 128 +Mmax = + I 3 R= SW) owl? 128 @ (V = 0) ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 STEEL AVERAGE WEB SHEAR STRESS f\= V|. fy = NU (detault) dty A Stor TEL other sections GENERAL SHEAR STRESS VQ . fe: oa SHEAR FLOW _ VQ Ae BOLT'S SPACING RT VQ SHEARING RESISTANCE R = Fy (eotrs) A bolt Noort FLEXURAL /BENDING STRESS Mc M eT x Bending Stress due +o Cwature MODULAR RATIO ‘F- modulus of elashcity f, = Ec | . @ radivs Of wrvature n= Es C | mpration Aber te NA. Ec Radius Of gyration (r) ALLOWABLE STRESS OF CONCRETE in fz IN STEEL DESIGN A Fe= O45 fc’ Moment of inevna (Trans fer F) Note: [ concrete stress = palaki | Tua = Tee + Ad’ weak — trong = paliit Secrion Moduus ASD = no factored load $= Tr Uniform Load (w) C w= Pressure * trilutary width | Reminder : elf Fy is NOT given, use A3G: Fy= 249 Mra 3 Fu = 400 MPa OR: Fy = 390 MPa; 414-420 MFa elf electrode number was given. mulh ply by 3.6 © Estee: = 200 GPa = 20010" MPa ENGR. ALEAH PEARL JAVIER ONCE, RCE 2024 
