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Axial Stress, Strain and Deformation Axial Stress Axial Strain Modulus of Elasticity Axial Deformation B cet pee galt oA Tis “€ AE stress 7 ductile materials yield strength ultimate strength JSracture point iat Plastic Vp elongation,6]| eat region vp v oKe g PL strain =eL,=—(L,) =—2 . Hooke’s law—+o = Ee 6 = eho ES 0 AE #1. Problem A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the required thickness of the tube if the stress is limited to 120 MPa- N : =) — Se fai pet A= (a? =?) me ( di la, A o 4 400,000 1 —__ = -(¢@2 - 2 720 4 do 1007) d, = 119.35 mm =d, 119.35 — 100 y SE im) T = 400 kN #2. PROBLEM The concrete post as shown in the figure is reinforced axially with four symmetrically placed steel bars, each of cross-sectional area 900 mm?. Compute the stress in each material when the 1000- KN axial load is applied. The moduli of elasticity are 200 GPa for steel and 14 GPa for concrete. A, = 3,600 mm? A, = 86,400 mm? | Pk Pt i P=P.+P, a ae -—5 P. = 626.87 kN .° P, = 1,000 — 626.87 = 373.13 kN =e= Asks i [= ep = es | iR=P, ASE. AE E. ‘For concrete: i P, _ 626.86x103 PL H H = —=—_—_—_ = [7.26 MP. €=7, (P=R+R % =A, 86,400 AE AsEs Heated Fe AcEc For steel: ; ° i P, _ 373.14x10° 3,600(200 Lee) 2 Tae ‘1,000 =p. +P. Foo] 6 = 7-= 3699 — ~03.65 MPa 86,400(14)| | eI- Flexoea tin. (Ciesla Seyi orig. = trans. #2. PROBLEM The concrete post as shown in the figure is reinforced > axially with four symmetrically placed steel bars, each of cross-sectional area 900 mm?. Compute the stress in each Ge) = =) material when the 1000-kN axial load is applied. The AE orig. Trane: moduli of elasticity are 200 GPa for steel and 14 GPa for Kral ea. Comite: Ac = 86,400mm? axial we Tk Berens. Vv Vv in For concrete: “Reed ELE Te teal) PL woot A 3 400( A A-E, = AsEs ENE ere 1 a (GB) 1,000x10° ec ea 1 ee 86,400 + 3,600 (FP) =(7.26 mPa) modular ratio, 1 #4, Problem A load P=100kN is supported by three (3) 20mm diameter wires as shown. Determine: 1. The force in wire AD. 2. The force in wire BD. 3. The vertical deflection of point D. oa xR =0 AB ansnlewnne. f 8 4 = 31.62 kN ” gla +Tp = 100 — eq. 1 H | (Tp = 49.41 kN) From compatibility of deformations: 4 i fA = —> Se, —4ep = 0 49.41x10° (4,000) Dee eo = "27 (200,000) : T4(5) = Ta(4)] _ ‘ —— (200,000) D AE AE | EaT 25T, — 167, = 0 —>eq.2 P = 100 kN #5. Problem Figure STR-001 shows a steel, aluminum and bronze connected to each other. P=25 kN. Aluminum i t fencion ‘A=400 mm? Bronze P _ 50,000 = Comes 1. Spronze = 7 = A 200 — 250 =(250 MPa) some) A- 22) PL _ 125,000(2,500) | = t t 2. Bsceet = FE = “590(200,000) A i H ! 1 =(.125 mm. \ | 1 1 i 4 3 o P 25,000 1 1 . Ealuminum = GF = arp aansco nn E AE 400(69,000 ! QS kN | 25 kN sara ( ») > ea 009%mm/mm i) 1 125k, FOO Kg] SEN Pe Te ey imple Shear Stress and Shear Strain Simple Shear Stress, t VIA, Simple Shear Strain, | Ly Modulus of Rigidity, G=1y eo oe A punch for making holes in steel plates is shown in Fig. STR-021. Assume that a punch having a diameter of 19 mm is used to punch a hole in a 6-mm plate, as shown in the cross-sectional view. Force P =125 kN. 1, Assuming that the tip of the punch is flat, determine the bearing stress between the punch and the steel plate. F 125,000 P=125kN oa =|440.87 MPa 2. What is the shear stress in the plate? _V_ 125,000 A 1(19)(6) = (349.02 MPa Problem 2. A steel strut S serving as a brace for a boat hoist transmits a compressive force P = 54kN to the deck of a pier as shown in figure STR-022. The strut has a hollow square cross section with a wall thickness t =12mm and the angle 8 between the strut and the horizontal is 40°. A pin through the strut transmits the compressive force from the strut to two gussets G that are welded to the base plate 8. Four anch Its fasten the base plate to the deck. The diameter of the pin is d,j,=18mm, the thickness of the gussets ea the thickness of the base plate is t, = 8mm, and the diameter of the anchor bolts is d,,.4 =12mm. Disregarding any friction between the base plate and the deck P/2 1. Determine the shear stress in the pin. 2. Determine the shear stress in the anchor bolts. & (Su, 3. Determine the bearing stress in the gussets. es = = Projected bearing area a F 54,000(0.5 ee ei 0 = 5 edgy ~ LOOM ANCHOR BOLT / [ Mathematically, J =/, + ly For solid shaft: For hollow shaft: Problem 3. A solid steel shaft 60 mm in diameter and 6.5 m long is subject to torques as shown in the figure. The shaft is attached to a rigid support at the left end. The modulus of rigidity of steel is 83 GPa. 1. Find the reaction at the rigid support. T 2. Calculate the max. shearing stress. — 3. Determine the total angle of twist at 4 1200 Nm 1000 N: m 800 N- m the free end of the shaft. Tr f : i F Ze — 1000 i 800 i Sar 5 ». T=0] H ig i —200 4 i —T — 1,200 + 1,000 —- 800 =0 . , T =-1,000N-—m Tr 7(3) 16T max Oe via apt ~ TDs _ 16(1,000x10%) _ Tmax =—s eye = 358 MPa Problem 3. A solid steel shaft 60 mm in diameter and 6.5 m long is subject to torques as shown in the figure. The shaft is attached to a rigid support at the left end. The modulus of rigidity of steel is 83 GPa. y om BA 4s m—-——— 3m “| 1. Find the reaction at the rigid support. T 2. Calculate the max. shearing stress. 3. Determine the total angle of twist at 4 1200 Nm 1000 Nem 800 N- m the free end of the shaft. 4 : i F 1,000 : 800 H —+ _ i Yr =o) 10 =200 —T — 1,200 + 1,000 —- 800 =0 4 23 TL T =-1,000N-—m total ~ JG _ # _ [1,000(2) — 200(1.5) + 800(3)](1x10%) 35 (604)(83,000) =(00388 rad. or 222 deg} Problem 4. A compound shaft ABC is attached to rigid supports at A and C. A torque T is applied at B. The ff. data are given: 7 — Te If T, = 1.963 i pF—+Tp =? a = ae B Tpy = 2.522, 0K BL ip 167 m7 OS de Tp = Tq +Te = 1.963 + 2.522 60 = Ter (37.5) 167, =(4.485 kN — m) Z 75*— 50%) 0 = =o ine | Thy = 3.988 kN —m Ay = 1963 kN = 7) Be = FA ~~ JG} a6 b -0~ y [2 4 1.963 x10(1.5x103) ‘Bye SUE lye BQ 35 (50*)(83,000) . =T,(2) | T (1.5) | TT a O0e7a LETS 33 (75* — 50*)(35,000)| |] x5 (50*) (83,000) i T, is = 1.285 = Thy = 1.2857, = 2.522A C x Problem 4. A compound shaft ABC is attached to rigid supports at A and C. A torque T is applied at B. The ff. data are given: Tt If T, = 1.963 Typ = 2.522, 0K Tp = T+ Te = 1.963 +2.522 =(4.485 kN — m) mee ‘BC IG Ls 1.963 x106 (1.5x103) 35 (50*)(83,000) =(0.0578 rad. or 3.313 deg. Ty =? ie A B c Tr _ 167 t= J ce max fr a y J ~~ 2.522% 10°(25) we oP ca ee BE: min 35(754 ~ 504) = 25.29 MPa Flexural Stress Mc Flexure Formula: tb ——_— vf where f, = flexural stress or bending stress M = bending moment at section listance from NA to the point where = + f, is being computed. For max f,,, ¢ is the distance from NA to extreme fiber of the section. I = moment of inertia of section Beam The term beam refers to a slender bar that carries transverse loading; that is, the applied forces are perpendicular to the bar. pa mut Lt Vv Positive bending moment tends to compress the upper part of the beam and elongate the lower part. Negative bending moment tends to elongate the upper part of the beam and compress the lower part of the beam. Positive shear force tends to rotate the material clockwise. Negative shear force tends to rotate the material counterclockwise. Problem 6 A built-up wooden beam section is made of two 60 mm x 120 mm sections glued on the lower sides of a 60 mm x 240 mm section as shown. The section is subject to a positive bending moment of 12 kN-m. ur Fojmax = Bee _ 12x109(90) foe 1. fomax = T73 33%106 pe Me a0 i =062 MPa _ a | 12x10°(150) Qu hbmnex = ene bmax = 179 32x10 pu = (16.03 MPal ° 60 60) GD (Np I = 112.32x10° mm* ten. wy ee 
