STRESS. Bar. ! Stress. ! Average Normal Stress in an Axially Loaded. ! Average Shear Stress. ! Allowable Stress. ! Design of Simple Connections
|
|
- Leo Sanders
- 6 years ago
- Views:
Transcription
1 STRESS! Stress Evisdom! verage Normal Stress in an xially Loaded ar! verage Shear Stress! llowable Stress! Design of Simple onnections 1
2 Equilibrium of a Deformable ody ody Force w F R x w(s). D s y Support Reaction D y
3 Table 1 Supports for oplanar Structures Type of onnection Idealized Symbol Reaction Number of Unknowns (1) θ Light cable θ θ One unknown. The reaction is a force that acts in the direction of the cable or link. () Rollers and rockers F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. (3) Smooth surface F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. 3
4 Type of onnection Idealized Symbol Reaction Number of Unknowns (4) Smooth pin or hinge F y F x Two unknowns. The reactions are two force components. (5) F x F y Two unknowns. The reactions are a force and moment. Internal pin (6) Fixed support F x M F y Three unknowns. The reactions are the moment and the two force components. 4
5 Equation of Equilibrium Σ F 0 Σ M O 0 Σ F x 0 Σ F y 0 Σ F z 0 Σ M x 0 Σ M y 0 Σ M z 0 5
6 Internal Resultant Loadings F section F 3 F 1 F 4 y F N M O x M O N V F 3 F 1 V F 4 6
7 Example 1 Determine the resultant internal loadings acting on the cross section at of the beam shown. 70 N/m 3 m 6 m 7
8 Support Reactions 70 N/m (1/)(9)(70) 115 N M 3645 N m 3 m 6 m 540 N Internal Loading y 115 N 180 N/m 540 N V 135 N 70 N/m 90 N/m M 3645 N m 3 m y 115 N -540() N m M 540 N N/m N 1080 N m N m 4 m 8
9 Shear and bending moment diagram (1/)(9)(70) 115 N 70 N/m 3645 N m 115 N V (N) 3 m 6 m M (N m) b Side View 540 N 1080 N m flanges t w web 540 N Front View h 9
10 Example Determine the resultant internal loadings acting on the cross section at of the machine shaft shown. The shaft is supported by bearings at and, which exert only vertical forces on the shaft. 00 mm 5 N 800 N/m D 100 mm 100 mm 50 mm 50 mm 10
11 Support Reactions (800 N/m)(0.15 m) 10 N/m 5 N x D 0.75 m m y y + Σ M 0: -(10)(0.75) + y (0.4) - (5)(0.5) 0, y N + Σ F x 0: x 0 x 0, + Σ F y 0: y y N, Internal Loading 800(0.05) 40 N + ΣF x 0: N 0 0. m N 0.05 m M V N + ΣF y 0: + ΣM 0: V 0, V N 18.75(0.5) + 40(0.05) + M 0 M N m 11
12 Internal Loading 800(0.05) 40 N 0. m M N m N 0 V N N 0.05 m 800(0.05) 40 N 0. m 5.69 N m N N 0.05 m 1
13 Example 3 The hoist consists of the beam and attached pulleys, the cable, and the motor. Determine the resultant internal loadings acting on the cross section at if the motor is lifting the kn load W with constant velocity. Neglect the weight of the pulleys and beam m 1 m m 0.5 m 0.15 m D W 13
14 0.15 m 1 m m 0.5 m 0.15 m D W W kn 0.15 m 1 m T kn N M V + + Σ F x 0: Σ F y 0: + N 0 N - kn - - V 0, V - kn W kn + Σ M 0: M - (0.15) + (1.15) 0 M - kn m 14
15 Example 4 Determine the resultant internal loadings acting on the cross section at G of the wooden beam shown. ssume the joints at,,, D, and E are pin-connected. 6.5 kn G D E 1 m 4 kn/m 0.6 m 0.6 m m 15
16 Support Reactions F Two - force 6.5 kn G D E E y E x 1 m 4 kn/m (1/)()(4) 4 kn 0.6 m 0.6 m /3 m (/3)() 4/3 m + Σ M E 0: 4(4/3) + 6.5(3.) - F (1) 0, F 6.1 kn + Σ F x 0: E x 0 E x -6.1 kn, + Σ F y 0: E y 0 E y 10.5 kn 16
17 Internal loadings acting on the cross section at G F 6.13 kn 34.0 kn 6.5 kn G 0.6 m Member G F V F o o 34.0 kn M N F D kn + ΣM G 0: M G sin39.81 o (0.6) + 6.5(0.6) 0 M G 9.17 kn m + + Joint ΣF x 0: ΣF y 0: -F sin50.19 o F 34.0 kn, (T) -F cos50.19 o -F D 0 F D -1.8 kn, () + ΣF x 0: 34.0 cos39.81 o + N G 0 N G -6.1 kn + ΣF y 0: sin39.81 o - V G 0 V G 15.3 kn 17
18 Example 5 Determine the resultant internal loadings acting on the cross section at of the pipe shown. The pipe has a mass of kg/m and is subjected to both a vertical force of 50 N and a couple moment of 70 N m at its end. It is fixed to the wall at. z x 50 N 70 N m 0.75 m 0.5 m D y 1.5 m 18
19 Free-ody Diagram z (M ) z (M ) y (M ) x x 50 N 70 N m (F ) x (F ) y 0.65 m (F ) z W D 9.81 N 0.5 m W D D 0.65 m y W D ( kg/m)(0.5 m)(9.81 N/kg) 9.81 N W D ( kg/m)(1.5 m)(9.81 N/kg) 4.5 N Equilibrium of Equilibrium Σ F x 0: (F ) x 0 Σ F y 0: (F ) y 0 Σ F z 0: (F ) z , (F ) z 84.3 N Σ (M ) x 0: (M ) x (0.5) (0.5) (0.5) 0, (M ) x N m Σ (M ) y 0: (M ) y (0.65) + 50(1.5) 0, (M ) y N m Σ (M ) x 0: (M ) z 0 19
20 Free-ody Diagram (F ) x 0 (F ) y 0 (F ) z 84.3 N (M ) x N m (M ) y N m (M ) z 0 z 77.8 N m 84.3 N 9.81 N 0.5 m 30.3 N m 4.5 N D x 50 N 0.65 m y 70 N m 0.65 m 0
21 Stress z +z Face +y Face y x +x Face 1
22 z σ z y compatibility, τ zx τ zy σ' x -x Face σ x σ' x σ y σ' y τ' xy τ yz σ z σ z -y Face x σ' y τ' yz τ' yx τ xz σ x τ' zy -z Face τ' xz τ xy τ' zx σ' z σ' y x z τ yx τ' yx x z σ y y τ' xy y z Top View τ xy τ' xy τ yx τ' yx τ zx τ xz σ' x y z τ yx x z y σ y x z x σ x y z τ xy y z
23 σ' y x z τ' O yx x z τ' xy y z O σ' x y z τ yx x z y σ y x z x σ x y z τ xy y z + ΣF y 0: - σ' y x z + σ y x z 0 σ' y σ y + ΣF x 0: σ x y z - σ' x y z 0 σ x σ' x + ΣM O 0: (τ xy y z)( x) - (τ yx x z)( y) 0 τ xy τ yx 3
24 z σ τ x y z τ zy x y zx x y z σ z x y z τ xz y z τ xy y z σ x y z τ yz x z σ y x z τ yx x z y x σ' y x z σ x y z σ' x y z σ y x z y x y x σ' z x y τ' yz x z z τ zy x y τ yz x z y τ' yx x z z τ' xy y z τ yx x z y τ xz y z z τ zx x y τ' xz y z y x τ' zy x y x τ xy y z x τ' zx x y 4
25 verage Normal Stress in an xially Loaded ar ssumptions The material must be - Homogeneous material - Isotropic material Internal force ross-sectional area External force 5
26 σ z F σ σ x y σ σ x y + F Rz ΣF ; z df σd σ σ 6
27 Example 6 The bar shown has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown. 1 kn 9 kn 4 kn D kn 35 mm 9 kn 4 kn 7
28 1 kn 9 kn 4 kn D 9 kn 4 kn 35 mm (kn) kn 1 30 x 10 mm 30 kn 35 mm σ σ σ 30 kn (0.035 m)(0.01 max m) 85.7 Ma 8
29 1 kn 9 kn 4 kn D 9 kn 4 kn 35 mm σ (Ma) kn x 10 mm 30 kn 35 mm σ σ σ 30 kn (0.035 m)(0.01 max m) 85.7 Ma 9
30 Example 7 The 80 kg lamp is supported b two rods and as shown. If has a diameter of 10 mm and has a diameter of 8 mm, determine which rod is subjected to the greater average normal stress. 60 o
31 Internal Loading y F 60 o F x verage Normal Stress d 10 mm o d 8 mm ΣF + + ΣF y x 80(9.81) N 4 0 ; F ( ) cos 60 o F o 0 ; F ( ) + F sin F 395. N, F 63.4 N σ σ F 395. N π (0.004 m) F 63.4 π (0.005 N m) Ma Ma The average normal stress distribution acting over a cross section of rod Ma 8.05 Ma 8.05 Ma 63.4 N 31
32 Example 8 Member shown is subjected to a vertical force of 3 kn. Determine the position x of this force so that the average compressive stress at is equal to the average tensile stress in the tie rod. The rod has a cross-sectional area of 400 mm and the contact area at is 650 mm. x 3 kn 00 mm 3
33 F x 3 kn 00 mm Internal Loading + ΣFy 0; F + F 3 0 (1) F + Σ M 0: -3(x) + F (00) () verage Normal Stress σ F F F 1.65F (3) Substituting (3) into (1), solving for F, the solving for F, we obtain F kn F kn The position of the applied load is determined from (); x 14 mm 33
34 onnections Simple Shear D τ avg V V V V τ avg 34
35 Single Shear Top View Side View Top View Side View V V 35
36 Double Shear / / / / Top View Top View Side View Side View V / V / V / V / 36
37 Normal Stress: ompression and Tension Load Tension Load a b a b d t b b bt t a a bt (b-d)t d b b ( b d) t Section a-a b t Section b-b 37
38 earing Stress b t d b b / / t d earing Stress bearing dt σ bearing dt 38
39 σ normal ( b d) t t θ dθ d b σ bearing 90 + ΣF x 0: d σ b( ) t cosθdθ 0 90 dt d 90 ( ) tσ b sinθdθ 90 σ bearing σ bearing td sin 90 td o * 39
40 xial Force Diagram for ompression Load on late b d t Tension - - ompression Normal Stress: σ, ( bd) t earing Stress σ bearing dt compression 40
41 xial Force Diagram for Tension Load on late a b d t Tension ompression Normal Stress: max + σ, ( b d) t earing Stress σ bearing dt Shearing Stress τ at tension t a τ / τ / 41
42 Shearing Stress on pin Single Shear τ 4
43 Double Shear / / / / / / / / / / / / / / τ τ / / 43
44 ure Shear z Section plane τ zy z τ yz y x y τ yzτzy x ure shear τ zy τ yz 44
45 Example 9 The bar shown a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar s cross-sectional area, determine the average normal stress and average shear stress acting on the material along (a) section plane a-a and (b) section plane b-b. 800 N a a b b 0 mm 30 o 0 mm 45
46 40 mm a b 0 mm 800 N a b 30 o 0 mm (a) section plane a-a a 800 N N 800 N a σ N ( N m)(0.04 m) 500 ka τ avg ka 46
47 40 mm a b 0 mm 800 N (b) section plane b-b a b 30 o 0 mm 800 N x 30 o V 800 cos30 o 69.8 N b b 30 o 30 o 800 N N 800sin30 o 400 N 40 mm o 40 sin 30 ; x x 80 mm σ State of b-b b τ avg σ b ka 15 ka τ avg N 400 N 15 ka (0.04 m)(0.08 m) V 69.8 N ka (0.04 m)(0.08 m) 47
48 Example 10 The wooden strut shown is suspended form a 10 mm diameter steel rod, which is fastened to the wall. If the strut supports a vertical load of 5 kn, compute the average shear stress in the rod at the wall and along the two shaded planes of the strut, one of which is indicated as abcd. b a c d 0 mm 40 mm 5 kn 48
49 b a c d 0 mm 40 mm verage shear stress in the rod at the wall 5 kn V 5 kn d 10 mm τ avg V 5 kn π (0.005 m) 5 kn 63.7 Ma 5 kn 63.7 Ma verage shear stress along the two shaded plane V.5 kn b a 5 kn V.5 kn c d τ avg V.5 kn (0.04 m)(0.0 m) 3.1 Ma 5 kn 3.1 Ma 49
50 Example 11 The inclined member show is subjected to a compressive force of 3 kn. Determine the average compressive stress along the areas of contact defined by and, and the average shear stress along the horizontal plane defined by ED kn 4 5 mm 40 mm 80 mm 50 mm 50
51 The average compressive stress along the areas of contact defined by and : 3 kn mm kn 4 40 mm 80 mm V 1.8 kn 50 mm F 3(3/5) 1.8 kn F 3(4/5).4 kn kn 4 σ F 1.8 kn (0.05 m)(0.040 m) 1800 ka 1.8 Ma 1.8 Ma σ F.4 kn (0.050 m)(0.040 m) 100 ka 1. Ma 1. Ma 51
52 The average shear stress along the horizontal plane defined by ED : 3 kn kn 4 5 mm 40 mm 80 mm 50 mm F 3(3/5) 1.8 kn V 1.8 kn F 3(4/5).4 kn τ avg V 1.8 kn (0.04 m)( Ma m) 56.5 ka 1.8 kn 0.56 Ma 5
53 llowable Stress F. S fail allow F. S σ σ fail allow F. S τ τ fail allow 6. Design of Simple onnections σ allow V τ allow ross-sectional rea of a Tension Member a τ allow a σ allow 53
54 ross-sectional rea of a onnector Subjected to Shear V τ allow σ allow Required rea to Resist earing (σ b ) allow (σ ) b allow 54
55 Required rea to Resist Shear by xial Load Uniform shear stress (τ allow ) d l τ allow πd 55
56 Example 1a The control arm is subjected to the loading shown. (a) Determine the shear stress for the steel at all pin (b) Determine normal stress in rod, plate D and E. The thickness of plate D and E is 10 mm. φ 10 mm φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn pin φ 0 mm 13 kn 75 mm 50 mm kn 56
57 F 00 mm 00 mm 13 kn 75 mm 50 mm kn x y 13 kn 75 mm 50 mm o 5 kn y Reaction + ΣM 0: F ( 0.) + 13(0.075) 5sin 36.87(0.15) 0 F 4.5 kn, + ΣF x 0: x + 5cos36.87 o 0 x 15.5 kn, x ΣF y 0: + y sin36.87 o 0 (15.5) + () y kn, kn 57
58 φ 35 mm (a) Shear stress 4.5 kn 00 mm 00 mm φ 5 mm φ 5 mm D E 5 3 pin φ 0 mm 13 kn 4 5 kn 75 mm 50 mm kn 13 kn 75 mm 50 mm 5 kn in (Double shear) in D and E (Single shear) kn kn τ D VD 13 kn 6.48 Ma ( π / 4)(0.05 ) D τ kn V kn 4.88 Ma ( π / 4)(0.0) τ E VE 5 kn Ma ( π / 4)(0.05 ) E 58
59 φ rod 10 mm (b) Normal stress 4.5 kn φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn kn 13 kn 75 mm 50 mm 5 kn σ ale σ 4.5 kn 56.7 Ma ( π / 4)(0.010) late D 13 kn 13 kn σ D 13 kn t 0.01 m 0.05 m σ D 13 kn (0.05)( 0.01) 6 Ma 59
60 φ rod 35 mm 4.5 kn φ 5 mm D E φ 5 mm 00 mm D E 50 mm 50 mm 13 kn 5 kn kn 13 kn 75 mm 50 mm 5 kn late E 5 kn 50 mm σ E 50 mm 5 kn 5 kn 0.05 m t 0.01 m 0.05 m σ E 5 kn ( )(0.01) 100 Ma 60
61 Example 1b The control arm is subjected to the loading shown. Determine the required diameter of the steel pin at if the allowable shear stress for the steel is τ allow 55 Ma. Note in the figure that the pin is subjected to double shear. 00 mm 13 kn 75 mm 50 mm kn 61
62 F 00 mm 00 mm 13 kn 75 mm 50 mm kn x y 13 kn 75 mm 50 mm kn y Internal Shear Force + ΣM 0: F ( 0.) 13(0.075) sin 36.87(0.15) 0 F kn + ΣF x 0: x + cos36.87 o 0 x 4.47 kn x ΣF y 0: + y sin36.87 o 0 (4.47) + (6.) y 6. kn 6.58 kn 6
63 13.15 kn 00 mm 00 mm 13 kn 75 mm 50 mm kn 6.58 kn 13 kn 75 mm 50 mm kn Required rea 6.58 kn τ V allow m 4 mm 13.9 kn 13.9 kn π d ( ) 4 mm d mm Use d 0 mm 63
64 Example 13a The two members are pinned together at as shown below. Top views of the pin connections at and are also given. If the pins have an allowable shear stress of τ allow 86 Ma, the allowable tensile stress of rod is (σ t ) allow 11 Ma and the allowable bearing stress is (σ b ) allow 150 Ma, determine to the smallest diameter of pins and,the diameter of rod and the thickness of necessary to support the load. Top view 13 kn m 0.6 m 64
65 Internal Force y 13 kn F x o 1 m 0.6 m + Σ M 0: 13 (1) + F sin 36.87(1.6) 0 F kn + Σ F y 0: y sin y 4.88 kn + Σ F x 0: x cos x kn 65
66 Diameter of the ins kn y 4.88 kn 13 kn F kn o x kn Top view 1 m 0.6 m in kn in F kn V 5.94 kn V 5.94 kn V kn V τ π 4 allow ( d ) 5.94 kn kn/ m mm mm 3 d 9.38 mm, Use d 10 mm π ( 4 V τ d allow ) kn kn / m mm mm 3 d mm, Use d 15 mm 66
67 Diameter of Rod kn y 4.88 kn 13 kn o F kn x kn 1 m 0.6 m kn kn ( σ t ) allow kn 10 kn/ m 10.9 mm 3 π 4 ( d ) 10.9 mm d 1.4 mm, Use d 15 mm 67
68 Thickness kn y 4.88 kn 13 kn F kn o x kn 1 m 0.6 m φ 10 mm t 0 kn φ 15 mm ( σ ) t b allow (0.010 ) t m kn kn t ( σ b ) allow (0.015) t 3 (σ b ) (σ b ) t m y comparison use t 8 mm 68
69 Example 13b The two members are pinned together at as shown below. Top views of the pin connections at and are also given. If the pins have an allowable shear stress of τ allow 86 Ma, the allowable tensile stress of rod is (σ t ) allow 11 Ma and the allowable bearing stress is (σ b ) allow 150 Ma, determine to the maximum load that the beam can supported. Top view φ 10 mm t 8 mm φ 15 mm m 0.6 m φ 15 mm t 8 mm 69
70 Internal Force y F x o 1 m 0.6 m + Σ M 0: ( 1) + F sin 36.87(1.6) 0 + ΣF 0 ; sin y y F 1.04 (T) y ΣF x 0 ; cos x x kn 70
71 Top view in (double shear) y F o 1 m 0.6 m φ 15 mm φ 10 mm in (single shear) F 1.04 V τ allow V V ka ( π / 4)(0.01) kn V 1.04 τ allow V ka ( π / 4)(0.015 ) kn 71
72 0.914 y o F m 0.6 m φ rod 15 mm Rod σ allow ka ( π / 4)(0.015 ) 19 kn 7
73 0.914 y F o Top view φ 10 mm t 10 mm ( σ b ) allow (0.010 )(0.010 ) kn φ 10 mm t 10 mm 1 m 0.6 m 0 kn F 1.04 (σ b ) (σ b ) φ 15 mm t 10 mm ( σ ) b allow (0.015 )( 0.01) kn t 10 mm φ 15 mm y comparison all kn 73
74 Example 14 The suspender rod is supported at its end by a fixed-connected circular disk as shown. If the rod passes through a 40-mm diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 0 kn load. The allowable normal stress for the rod is σ allow 60 Ma, and the allowable shear stress for the disk is τ allow 35 Ma. 40 mm t 40 mm τ allow d 0 kn 0 kn 74
75 40 mm t 40 mm τ allow d 0 kn 0 kn Diameter of Rod Thickness of Disk σ allow kn kn/ m τ V allow kn kn/ m π 0 d kn kn/ m π (0.0 m) t kn kn/ m d m 0.6 mm t 4.55x10-3 m 4.55 mm 75
76 Example 15 n axial load on the shaft shown is resisted by the collar at, which is attached to the shaft and located on the right side of the bearing at. Determine the largest value of for the two axial forces at E and F so that the stress in the collar does not exceed an allowable bearing stress at of (σ b ) allow 75 Ma and allowable shearing stress of the adhesive at of τ allow 100 Ma, and the average normal stress in the shaft does not exceed an allowable tensile stress of (σ t ) allow 55 Ma. F E 60 mm 0 mm 80 mm 76
77 F E 60 mm 0 mm 3 80 mm Load 3 osition xial Stress earing Stress (σ t ) allow kn/ m 3 π ( kn m) 3 3 (σ b ) allow kn/ m [ π (0.04 m) π (0.03 m) ] 55 kn 60 mm 80 mm 77
78 F E 60 mm 0 mm 3 80 mm Load 100 Shearing Stress 60 mm 3 τ 10 allow 3 kn/ m shear 3 osition 80 mm 0 mm 3 [π (0.04 m)(. 00 )] xial Stress kn earing Stress 55 kn Shearing Stress 3 55 kn 3 55 kn The largest load that can be applied to the shaft is 51.8 kn. 78
79 Example 16 The rigid bar shown supported by a steel rod having a diameter of 0 mm and an aluminum block having a cross-sectional area of 1800 mm. The18-mmdiameter pins at and are subjected to single shear. If the failure stress for the steel and aluminum is (σ st ) fail 680 Mpa and (σ al ) fail 70 Ma, respectively, and the failure shear stress for each pin is τ fail 900 Ma, determine the largest load that can be applied to the bar. pply a factor of safety of F.S.0. Steel 0.75 m luminum m 79
80 Steel 0.75 m F luminum 0.75 m 1.5 m m F + ΣM 0: (1.5 m) - F ( m) > F ΣM 0: (0.75 m) - F ( m) > F Rod ( σ st ) allow ( σ st ) F. S fail F ka 0.65 π( kn m) lock 70 ( σ al ) allow ( σ al ) F. S fail 3 10 ka kn 6 F m 80
81 Steel 0.75 m F 0.65 luminum 0.75 m 1.5 m m F in or τ allow τ fail F. S F pin ka 0.65 π(0.009 m) kn Summary Rod kn lock 168 kn in or kn Largest load 168 kn 81
Determine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
More information[5] Stress and Strain
[5] Stress and Strain Page 1 of 34 [5] Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law
More informationMECE 3321 MECHANICS OF SOLIDS CHAPTER 1
MECE 3321 MECHANICS O SOLIDS CHAPTER 1 Samantha Ramirez, MSE WHAT IS MECHANICS O MATERIALS? Rigid Bodies Statics Dynamics Mechanics Deformable Bodies Solids/Mech. Of Materials luids 1 WHAT IS MECHANICS
More informationMECHANICS OF MATERIALS. Prepared by Engr. John Paul Timola
MECHANICS OF MATERIALS Prepared by Engr. John Paul Timola Mechanics of materials branch of mechanics that studies the internal effects of stress and strain in a solid body. stress is associated with the
More informationChapter 1 Introduction- Concept of Stress
hapter 1 Introduction- oncept of Stress INTRODUTION Review of Statics xial Stress earing Stress Torsional Stress 14 6 ending Stress W W L Introduction 1-1 Shear Stress W W Stress and Strain L y y τ xy
More informationand F NAME: ME rd Sample Final Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points)
ME 270 3 rd Sample inal Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points) IND: In your own words, please state Newton s Laws: 1 st Law = 2 nd Law = 3 rd Law = PROBLEM
More informationSamantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2
Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force
More informationStatics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3
Statics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3 2 3 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the
More informationEquilibrium of a Particle
ME 108 - Statics Equilibrium of a Particle Chapter 3 Applications For a spool of given weight, what are the forces in cables AB and AC? Applications For a given weight of the lights, what are the forces
More informationProblem " Â F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn
Problem 0. Three cables are attached as shown. Determine the reactions in the supports. Assume R B as redundant. Also, L AD L CD cos 60 m m. uation of uilibrium: + " Â F y 0 ) R A cos 60 + R B + R C cos
More informationSample 5. Determine the tension in the cable and the horizontal and vertical components of reaction at the pin A. Neglect the size of the pulley.
Sample 1 The tongs are designed to handle hot steel tubes which are being heat-treated in an oil bath. For a 20 jaw opening, what is the minimum coefficient of static friction between the jaws and the
More informationModule 2 Stresses in machine elements. Version 2 ME, IIT Kharagpur
Module Stresses in machine elements Lesson Compound stresses in machine parts Instructional Objectives t the end of this lesson, the student should be able to understand Elements of force system at a beam
More informationMechanical Properties of Materials
Mechanical Properties of Materials Strains Material Model Stresses Learning objectives Understand the qualitative and quantitative description of mechanical properties of materials. Learn the logic of
More informationMECHANICS OF MATERIALS
Third E CHAPTER 1 Introduction MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University Concept of Stress Contents Concept of Stress
More informationMAAE 2202 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock final exam UNDER EXAM CONDITIONS. This means: Complete the exam in 3 hours. Work on your own. Keep your textbook closed. Attempt every question. After the
More informationAluminum shell. Brass core. 40 in
PROBLEM #1 (22 points) A solid brass core is connected to a hollow rod made of aluminum. Both are attached at each end to a rigid plate as shown in Fig. 1. The moduli of aluminum and brass are EA=11,000
More informationD : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each.
GTE 2016 Q. 1 Q. 9 carry one mark each. D : SOLID MECHNICS Q.1 single degree of freedom vibrating system has mass of 5 kg, stiffness of 500 N/m and damping coefficient of 100 N-s/m. To make the system
More informationD : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.
D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More information2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?
IDE 110 S08 Test 1 Name: 1. Determine the internal axial forces in segments (1), (2) and (3). (a) N 1 = kn (b) N 2 = kn (c) N 3 = kn 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at
More informationMECH 401 Mechanical Design Applications
MECH 401 Mechanical Design Applications Dr. M. O Malley Master Notes Spring 008 Dr. D. M. McStravick Rice University Updates HW 1 due Thursday (1-17-08) Last time Introduction Units Reliability engineering
More informationARC 341 Structural Analysis II. Lecture 10: MM1.3 MM1.13
ARC241 Structural Analysis I Lecture 10: MM1.3 MM1.13 MM1.4) Analysis and Design MM1.5) Axial Loading; Normal Stress MM1.6) Shearing Stress MM1.7) Bearing Stress in Connections MM1.9) Method of Problem
More information[8] Bending and Shear Loading of Beams
[8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight
More informationThe University of Melbourne Engineering Mechanics
The University of Melbourne 436-291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 9-22 from Hibbeler - Statics and Mechanics of Materials) A short
More informationTUTORIAL SHEET 1. magnitude of P and the values of ø and θ. Ans: ø =74 0 and θ= 53 0
TUTORIAL SHEET 1 1. The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. Knowing that the tension in the cable is 1349N, determine the moment
More informationStress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 4 ME 76 Spring 017-018 Dr./ Ahmed Mohamed Nagib Elmekawy Shear and Moment Diagrams Beam Sign Convention The positive directions are as follows: The internal shear force causes a
More informationPurpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on Exam 3.
ES230 STRENGTH OF MTERILS Exam 3 Study Guide Exam 3: Wednesday, March 8 th in-class Updated 3/3/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on
More informationReview Lecture. AE1108-II: Aerospace Mechanics of Materials. Dr. Calvin Rans Dr. Sofia Teixeira De Freitas
Review Lecture AE1108-II: Aerospace Mechanics of Materials Dr. Calvin Rans Dr. Sofia Teixeira De Freitas Aerospace Structures & Materials Faculty of Aerospace Engineering Analysis of an Engineering System
More informationSOLUTION 8 7. To hold lever: a+ M O = 0; F B (0.15) - 5 = 0; F B = N. Require = N N B = N 0.3. Lever,
8 3. If the coefficient of static friction at is m s = 0.4 and the collar at is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support
More informationNAME: Given Formulae: Law of Cosines: Law of Sines:
NME: Given Formulae: Law of Cosines: EXM 3 PST PROBLEMS (LESSONS 21 TO 28) 100 points Thursday, November 16, 2017, 7pm to 9:30, Room 200 You are allowed to use a calculator and drawing equipment, only.
More informationPROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D.
1. he uniform I-beam has a mass of 60 kg per meter of its length. Determine the tension in the two supporting cables and the reaction at D. (3/62) A( 500) m (5 23) m m = 60 kg/m determine the tension in
More informationSTATICS. Bodies. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Design of a support
4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies 2010 The McGraw-Hill Companies,
More informationStress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 3 ME 276 Spring 2017-2018 Dr./ Ahmed Mohamed Nagib Elmekawy Axial Stress 2 Beam under the action of two tensile forces 3 Beam under the action of two tensile forces 4 Shear Stress
More informationMechanics of Materials CIVL 3322 / MECH 3322
Mechanics of Materials CIVL 3322 / MECH 3322 2 3 4 5 6 7 8 9 10 A Quiz 11 A Quiz 12 A Quiz 13 A Quiz 14 A Quiz 15 A Quiz 16 In Statics, we spent most of our time looking at reactions at supports Two variations
More informationDESIGN OF BEAMS AND SHAFTS
DESIGN OF EAMS AND SHAFTS! asis for eam Design! Stress Variations Throughout a Prismatic eam! Design of pristmatic beams! Steel beams! Wooden beams! Design of Shaft! ombined bending! Torsion 1 asis for
More informationSamantha Ramirez, MSE
Samantha Ramirez, MSE Centroids The centroid of an area refers to the point that defines the geometric center for the area. In cases where the area has an axis of symmetry, the centroid will lie along
More informationME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam cross-sec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.
ME 323 - Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM-12:20PM Ghosh 2:30-3:20PM Gonzalez 12:30-1:20PM Zhao 4:30-5:20PM M (x) y 20 kip ft 0.2
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationMECHANICS OF MATERIALS
Fifth SI Edition CHTER 1 MECHNICS OF MTERILS Ferdinand. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Introduction Concept of Stress Lecture Notes: J. Walt Oler Teas Tech University Contents
More informationChapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using
More informationEquilibrium of a Rigid Body. Engineering Mechanics: Statics
Equilibrium of a Rigid Body Engineering Mechanics: Statics Chapter Objectives Revising equations of equilibrium of a rigid body in 2D and 3D for the general case. To introduce the concept of the free-body
More informationMechanical Engineering Ph.D. Preliminary Qualifying Examination Solid Mechanics February 25, 2002
student personal identification (ID) number on each sheet. Do not write your name on any sheet. #1. A homogeneous, isotropic, linear elastic bar has rectangular cross sectional area A, modulus of elasticity
More information3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM
3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity Two types of forces that act on it, the
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
More informationMechanics of Materials
Mechanics of Materials 2. Introduction Dr. Rami Zakaria References: 1. Engineering Mechanics: Statics, R.C. Hibbeler, 12 th ed, Pearson 2. Mechanics of Materials: R.C. Hibbeler, 9 th ed, Pearson 3. Mechanics
More informationCIVIL DEPARTMENT MECHANICS OF STRUCTURES- ASSIGNMENT NO 1. Brach: CE YEAR:
MECHANICS OF STRUCTURES- ASSIGNMENT NO 1 SEMESTER: V 1) Find the least moment of Inertia about the centroidal axes X-X and Y-Y of an unequal angle section 125 mm 75 mm 10 mm as shown in figure 2) Determine
More informationThree torques act on the shaft. Determine the internal torque at points A, B, C, and D.
... 7. Three torques act on the shaft. Determine the internal torque at points,, C, and D. Given: M 1 M M 3 300 Nm 400 Nm 00 Nm Solution: Section : x = 0; T M 1 M M 3 0 T M 1 M M 3 T 100.00 Nm Section
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More informationThe case where there is no net effect of the forces acting on a rigid body
The case where there is no net effect of the forces acting on a rigid body Outline: Introduction and Definition of Equilibrium Equilibrium in Two-Dimensions Special cases Equilibrium in Three-Dimensions
More informationQuestion 1. Ignore bottom surface. Solution: Design variables: X = (R, H) Objective function: maximize volume, πr 2 H OR Minimize, f(x) = πr 2 H
Question 1 (Problem 2.3 of rora s Introduction to Optimum Design): Design a beer mug, shown in fig, to hold as much beer as possible. The height and radius of the mug should be not more than 20 cm. The
More information5.2 Rigid Bodies and Two-Dimensional Force Systems
5.2 Rigid odies and Two-Dimensional Force Systems 5.2 Rigid odies and Two-Dimensional Force Systems Procedures and Strategies, page 1 of 1 Procedures and Strategies for Solving Problems Involving Equilibrium
More informationLevers. 558 A Textbook of Machine Design
558 A Textbook of Machine Design C H A P T E R 15 Levers 1. Introduction.. Application of Levers in Engineering Practice.. Design of a Lever. 4. Hand Lever. 5. Foot Lever. 6. Cranked Lever. 7. Lever for
More information2014 MECHANICS OF MATERIALS
R10 SET - 1 II. Tech I Semester Regular Examinations, March 2014 MEHNIS OF MTERILS (ivil Engineering) Time: 3 hours Max. Marks: 75 nswer any FIVE Questions ll Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~~~~
More information8.1 Internal Forces in Structural Members
8.1 Internal Forces in Structural Members 8.1 Internal Forces in Structural Members xample 1, page 1 of 4 1. etermine the normal force, shear force, and moment at sections passing through a) and b). 4
More informationChapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING )
Chapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING ) 5.1 DEFINITION A construction member is subjected to centric (axial) tension or compression if in any cross section the single distinct stress
More informationEngineering Mechanics: Statics
Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant
More informationPDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics
Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.
More informationCHAPTER 4 Stress Transformation
CHAPTER 4 Stress Transformation ANALYSIS OF STRESS For this topic, the stresses to be considered are not on the perpendicular and parallel planes only but also on other inclined planes. A P a a b b P z
More information(48) CHAPTER 3: TORSION
(48) CHAPTER 3: TORSION Introduction: In this chapter structural members and machine parts that are in torsion will be considered. More specifically, you will analyze the stresses and strains in members
More informationEng Sample Test 4
1. An adjustable tow bar connecting the tractor unit H with the landing gear J of a large aircraft is shown in the figure. Adjusting the height of the hook F at the end of the tow bar is accomplished by
More informationAnnouncements. Equilibrium of a Rigid Body
Announcements Equilibrium of a Rigid Body Today s Objectives Identify support reactions Draw a free body diagram Class Activities Applications Support reactions Free body diagrams Examples Engr221 Chapter
More informationCode No: R Set No. 1
Code No: R05010302 Set No. 1 I B.Tech Supplimentary Examinations, February 2008 ENGINEERING MECHANICS ( Common to Mechanical Engineering, Mechatronics, Metallurgy & Material Technology, Production Engineering,
More informationMECE 3321: Mechanics of Solids Chapter 6
MECE 3321: Mechanics of Solids Chapter 6 Samantha Ramirez Beams Beams are long straight members that carry loads perpendicular to their longitudinal axis Beams are classified by the way they are supported
More informationEngineering Science OUTCOME 1 - TUTORIAL 4 COLUMNS
Unit 2: Unit code: QCF Level: Credit value: 15 Engineering Science L/601/10 OUTCOME 1 - TUTORIAL COLUMNS 1. Be able to determine the behavioural characteristics of elements of static engineering systems
More informationStress Transformation Equations: u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa. we obtain, cos u + t xy sin 2u. s x = s x + s y.
014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. Determine the normal stress and shear stress acting
More informationBEAM A horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam
BEM horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam INTERNL FORCES IN BEM Whether or not a beam will break, depend on the internal resistances
More informationAnnouncements. Trusses Method of Joints
Announcements Mountain Dew is an herbal supplement Today s Objectives Define a simple truss Trusses Method of Joints Determine the forces in members of a simple truss Identify zero-force members Class
More informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion
EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 3 Torsion Introduction Stress and strain in components subjected to torque T Circular Cross-section shape Material Shaft design Non-circular
More informationMechanics of Solids I. Transverse Loading
Mechanics of Solids I Transverse Loading Introduction o Transverse loading applied to a beam results in normal and shearing stresses in transverse sections. o Distribution of normal and shearing stresses
More informationIDE 110 Mechanics of Materials Spring 2006 Final Examination FOR GRADING ONLY
Spring 2006 Final Examination STUDENT S NAME (please print) STUDENT S SIGNATURE STUDENT NUMBER IDE 110 CLASS SECTION INSTRUCTOR S NAME Do not turn this page until instructed to start. Write your name on
More informationUnit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir
Unit III Theory of columns 1 Unit III Theory of Columns References: Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength of Materials", Tata
More information7 TRANSVERSE SHEAR transverse shear stress longitudinal shear stresses
7 TRANSVERSE SHEAR Before we develop a relationship that describes the shear-stress distribution over the cross section of a beam, we will make some preliminary remarks regarding the way shear acts within
More information5. Plane Kinetics of Rigid Bodies
5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.6 Work and energy relations 5.7 Impulse
More informationQuestions from all units
Questions from all units S.NO 1. 1 UNT NO QUESTON Explain the concept of force and its characteristics. BLOOMS LEVEL LEVEL 2. 2 Explain different types of force systems with examples. Determine the magnitude
More informationME 202 STRENGTH OF MATERIALS SPRING 2014 HOMEWORK 4 SOLUTIONS
ÇANKAYA UNIVERSITY MECHANICAL ENGINEERING DEPARTMENT ME 202 STRENGTH OF MATERIALS SPRING 2014 Due Date: 1 ST Lecture Hour of Week 12 (02 May 2014) Quiz Date: 3 rd Lecture Hour of Week 12 (08 May 2014)
More informationSymmetric Bending of Beams
Symmetric Bending of Beams beam is any long structural member on which loads act perpendicular to the longitudinal axis. Learning objectives Understand the theory, its limitations and its applications
More informationAnnouncements. Equilibrium of a Particle in 2-D
nnouncements Equilibrium of a Particle in 2-D Today s Objectives Draw a free body diagram (FBD) pply equations of equilibrium to solve a 2-D problem Class ctivities pplications What, why, and how of a
More informationMECHANICS OF MATERIALS
CHTER MECHNICS OF MTERILS 10 Ferdinand. Beer E. Russell Johnston, Jr. Columns John T. DeWolf cture Notes: J. Walt Oler Texas Tech University 006 The McGraw-Hill Companies, Inc. ll rights reserved. Columns
More information7.5 Elastic Buckling Columns and Buckling
7.5 Elastic Buckling The initial theory of the buckling of columns was worked out by Euler in 1757, a nice example of a theory preceding the application, the application mainly being for the later invented
More informationIf the solution does not follow a logical thought process, it will be assumed in error.
Please indicate your group number (If applicable) Circle Your Instructor s Name and Section: MWF 8:30-9:20 AM Prof. Kai Ming Li MWF 2:30-3:20 PM Prof. Fabio Semperlotti MWF 9:30-10:20 AM Prof. Jim Jones
More informationNational Exams May 2015
National Exams May 2015 04-BS-6: Mechanics of Materials 3 hours duration Notes: If doubt exists as to the interpretation of any question, the candidate is urged to submit with the answer paper a clear
More informationSolution: T, A1, A2, A3, L1, L2, L3, E1, E2, E3, P are known Five equations in five unknowns, F1, F2, F3, ua and va
ME 323 Examination # 1 February 18, 2016 Name (Print) (Last) (First) Instructor PROBLEM #1 (20 points) A structure is constructed from members 1, 2 and 3, with these members made up of the same material
More information11.1 Virtual Work Procedures and Strategies, page 1 of 2
11.1 Virtual Work 11.1 Virtual Work rocedures and Strategies, page 1 of 2 rocedures and Strategies for Solving roblems Involving Virtual Work 1. Identify a single coordinate, q, that will completely define
More informationD e s i g n o f R i v e t e d J o i n t s, C o t t e r & K n u c k l e J o i n t s
D e s i g n o f R i v e t e d J o i n t s, C o t t e r & K n u c k l e J o i n t s 1. Design of various types of riveted joints under different static loading conditions, eccentrically loaded riveted joints.
More informationENGR-1100 Introduction to Engineering Analysis. Lecture 13
ENGR-1100 Introduction to Engineering Analysis Lecture 13 EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS Today s Objectives: Students will be able to: a) Identify support reactions, and, b) Draw a free-body
More informationModule 11 Design of Joints for Special Loading. Version 2 ME, IIT Kharagpur
Module 11 Design of Joints for Special Loading Version ME, IIT Kharagpur Lesson Design of Eccentrically Loaded Welded Joints Version ME, IIT Kharagpur Instructional Objectives: At the end of this lesson,
More informationThe centroid of an area is defined as the point at which (12-2) The distance from the centroid of a given area to a specified axis may be found by
Unit 12 Centroids Page 12-1 The centroid of an area is defined as the point at which (12-2) The distance from the centroid of a given area to a specified axis may be found by (12-5) For the area shown
More informationUNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich
UNIVERSITY OF SASKATCHEWAN ME 313.3 MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich A CLOSED BOOK EXAMINATION TIME: 3 HOURS For Marker s Use Only LAST NAME (printed): FIRST
More informationCOLUMNS: BUCKLING (DIFFERENT ENDS)
COLUMNS: BUCKLING (DIFFERENT ENDS) Buckling of Long Straight Columns Example 4 Slide No. 1 A simple pin-connected truss is loaded and supported as shown in Fig. 1. All members of the truss are WT10 43
More informationBPSC Main Exam 2019 ASSISTANT ENGINEER. Test 11. CIVIL ENGINEERING Subjective Paper-I. Detailed Solutions. Detailed Solutions
etailed Solutions SC Main Exam 19 SSISTNT ENGINEER CIVI ENGINEERING Subjective aper-i Test 11 Q.1 (a) Solution: (i) Calculation of maximum moment at 8 m. For maximum moment Case 1: Case : Case : Case 4:
More informationA concrete cylinder having a a diameter of of in. mm and elasticity. Stress and Strain: Stress and Strain: 0.
2011 earson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 8 1. 3 1. concrete cylinder having a a diameter of of 6.00
More informationTorsion of Shafts Learning objectives
Torsion of Shafts Shafts are structural members with length significantly greater than the largest cross-sectional dimension used in transmitting torque from one plane to another. Learning objectives Understand
More informationStrength of Material. Shear Strain. Dr. Attaullah Shah
Strength of Material Shear Strain Dr. Attaullah Shah Shear Strain TRIAXIAL DEFORMATION Poisson's Ratio Relationship Between E, G, and ν BIAXIAL DEFORMATION Bulk Modulus of Elasticity or Modulus of Volume
More informationI B.TECH EXAMINATIONS, JUNE ENGINEERING MECHANICS (COMMON TO CE, ME, CHEM, MCT, MMT, AE, AME, MIE, MIM)
Code.No: 09A1BS05 R09 SET-1 I B.TECH EXAMINATIONS, JUNE - 2011 ENGINEERING MECHANICS (COMMON TO CE, ME, CHEM, MCT, MMT, AE, AME, MIE, MIM) Time: 3 hours Max. Marks: 75 Answer any FIVE questions All questions
More informationName (Print) ME Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM
Name (Print) (Last) (First) Instructions: ME 323 - Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM Circle your lecturer s name and your class meeting time. Gonzalez Krousgrill
More informationQ. 1 Q. 5 carry one mark each.
General ptitude G Set-8 Q. 1 Q. 5 carry one mark each. Q.1 The chairman requested the aggrieved shareholders to him. () bare with () bore with (C) bear with (D) bare Q.2 Identify the correct spelling out
More information4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight.
1 1 wooden block of mass 0.60 kg is on a rough horizontal surface. force of 12 N is applied to the block and it accelerates at 4.0 m s 2. wooden block 4.0 m s 2 12 N hat is the magnitude of the frictional
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS
EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering
More information1-1 Locate the centroid of the plane area shown. 1-2 Determine the location of centroid of the composite area shown.
Chapter 1 Review of Mechanics of Materials 1-1 Locate the centroid of the plane area shown 650 mm 1000 mm 650 x 1- Determine the location of centroid of the composite area shown. 00 150 mm radius 00 mm
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 1 - STATIC AND DYNAMIC FORCES TUTORIAL 3 STRESS AND STRAIN
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 1 - STATIC AND DYNAMIC FORCES TUTORIAL 3 STRESS AND STRAIN 1 Static and dynamic forces Forces: definitions of: matter, mass, weight,
More information