1 COMBINED LOADS In many structures the members are required to resist more than one kind of loading (combined loading). These can often be analyzed by superimposing the stresses and strains cause by each load acting separately. Superposition of stresses and strains is permissible only under the following conditions: a.the stresses and the strains must be a linear function of the applied loads (Hooke s law must be obeyed and the displacements must be small). b.there must be no interaction between the various loads. Eamples: wide flange beam supported by a cable (combined bending and aial load), cylindrical pressure vessel supported as a beam, and shaft in combined torsion and bending.
2 Method of Analysis:.Select the point on the structure where the stresses and the strains are to be determined..for each load on the structure, determine the stress resultant at the cross section containing the selected point...calculate the normal and shear stresses at the selected point due to each of the stress resultant. 4.Combine the individual stresses to obtain the resultant stresses at the selected point. 5.Determine the principal stresses and maimum shear stresses at the selected point. 6.Determine the strains at the point with the aid of Hooke s law for plane stress. 7.Select additional points and repeat the process. P A VQ Ib Τρ Ι ρ pr t My I
3 Illustration of the Method: The bar shown is subjected to two types of loads: a torque T and a vertical load P. Let us select arbitrarily two points. Point A (top of the bar) and point B (side of the bar in the same cross section). The resulting stresses acting across the section are the following: A twisting moment equal to the torque T. A bending moment M equal to the load P times the distance b. A shear force V equals to the load P.
4 The twisting moment Τ produces a torsional shear stresses Tr T torsion I πr Polar The stress acts horizontally to the left at point A and vertically downwards at point B. The bending moment M produces a tensile stress at point A Mr bending I 4M πr However, the bending moment produces no stress at point B, because B is located on the neutral ais. The shear force V produces no shear stress at the top of the bar (point A), but at point B the shear stress is as follows: shear VQ Ib 4V A A and are acting in point A, while the and are acting in point B.
5 Note that the element is in plane stress with A, y 0, and y -. A stress element in point B is also in plane stress and the only stresses acting on this element are the shear stresses and. Therefore y 0 and y - ( + ). At point A: A, y 0, and y - At point B y 0 and y - ( + ).
6 Of interest are the points where the stresses calculated from the fleure and shear formulas have maimum or minimum values, called critical points. For instance, the normal stresses due to bending are largest at the cross section of maimum bending moment, which is at the support. Therefore, points C and D at the top and bottom of the beam at the fied ends are critical points where the stresses should be calculated. Selection of Critical Areas and Points If the objective of the analysis is to determine the largest stresses anywhere in the structure, then the critical points should be selected at cross sections where the stress resultants have their largest values. Furthermore, within those cross sections, the points should be selected where either the normal stresses or the shear stresses have their largest values.
7 Stress at which point?
8 The rotor shaft of an helicopter drives the rotor blades that provide the lifting force to support the helicopter in the air. As a consequence, the shaft is subjected to a combination of torsion and aial loading. For a 50mm diameter shaft transmitting a torque Τ.4kN.m and a tensile force P 5kN, determine the maimum tensile stress, maimum compressive stress, and maimum shear stress in the shaft. Solution The stresses in the rotor shaft are produced by the combined action of the aial force P and the torque Τ. Therefore the stresses at any point on the surface of the shaft consist of a tensile stress o and a shear stress o. The tensile stress P 5kN A π 4 The shear stress to is obtained from the torsion formula ( 0.05m) 6.66MPa Tr I (.4kN. m) 97. Torsion P ( 0.05) 78 4 π 0.05 MPa
9 Knowing the stresses o and o, we can now obtain the principal stresses and maimum shear stresses. The principal stresses are obtained from + y y, + ± ( ) y The maimum in-plane shear stresses are obtained using the formula, MPa 7MPa 0 + ± ( ) MPa y MAX y 0 ( 97.78) Because the principal stresses and have opposite signs, the maimum in-plane shear stresses are larger than the maimum out-of-plane shear stresses. Therefore, the maimum shear stress in the shaft is 0MPa. Will it fail if yield 480MPa? MSST SF 480MPa 0MPa. VM ( 5) ( 5)( 7) + ( 7) 480MPa DET SF 8.MPa.65 8.MPa
10 A thin wall cylindrical pressure vessel with a circular cross section is subjected to internal gas pressure p and simultaneously compressed by an aial load P k. The cylinder has inner radius r.in. And wall thickness t 0.5in. Determine the maimum allowable internal pressure p allow based upon an allowable shear stress of 6500psi in the wall of the vessel. Solution The stresses on the wall of the pressure vessel are caused by a combined action of the internal pressure and the aial force. We can isolate a stress element in point A. The -ais is parallel to the longitudinal ais of the pressure vessel and the y-ais is circumferential. Note that there are no shear stresses acting on the element. The longitudinal stress is equal to the tensile stress produced by the internal pressure minus the compressive stress produced by the aial force. pr t P A pr t P πrt
11 The circumferential stress y is equal to the tensile stress produced by the internal pressure. Note that y >. Since no shear stresses act on the element the above stresses are also the principal stresses substituting numerical values pr t pr t (. in) p 05. in P πrt 4.0 p p (. in) k ( 05. in) [ π (.in)( 0.5in) ] 7.0 p y 606 y pr t pr t pr t psi P πrt In-Plane Shear Stresses The maimum in-plane shear stress is ( ) (( 4.0 p) ( 7.0 p 606psi) ) Ma.5 p + 0 psi Since ma is limited to 6500psi then psi.4 p + 0 psi pallowed 990 psi
12 Out-of-Plane Shear Stresses The maimum out-of-plane shear stress is either From the first equation we get psi p allowed 70.5 p 0 psi psi Ma Ma From the second equation we get: psi 7.0 p pallowed 98psi Allowable internal pressure Comparing the three calculated values for the allowable pressure, we see that (p allow ) 98psi governs. At this pressure, the principal stresses are 000psi and 40psi. These stresses have the same signs, thus confirming that one of the outof-plane shear stresses must be the largest shear stress.
13 A sign of dimensions.0m.m is supported by a hollow circular pole having outer diameter 0mm and inner diameter 80mm (see figure). The sign offset 0.5m from the centerline of the pole and its lower edge is 6.0m above the ground. Determine the principal stresses and maimum shear stresses at points A and B at the base of the pole due to wind pressure of.0kpa against the sign. Solution Stress Resultant: The wind pressure against the sign produces a resultant force W that acts at the midpoint of the sign and it is equal to the pressure p times the area A over which it acts: (.0kPa)(.0m.m) 4. kn W pa 8 The line of action of this force is at height h 6.6m above the ground and at distance b.5m from the centerline of the pole. The wind force acting on the sign is statically equivalent to a lateral force W and a torque Τ acting on the pole.
14 The torque is equal to the force W times the distance b: T T Wb 7.kN ( 4.8kN )(.5m) m The stress resultant at the base of the pole consists of a bending moment M, a torque Τ and a shear force V. Their magnitudes are: M Wh (4.8kN)(6.6m).68kN.m Τ 7.kN.m V W 4.8kN Eamination of these stress resultants shows that maimum bending stresses occur at point A and maimum shear stresses at point B. Therefore, A and B are critical points where the stresses should be determined. Stresses at points A and B The bending moment M produces a tensile stress a at point A, but no stress at point B (which is located on the neutral ais)
15 d M (.68kN )( 0.m) a 54. 9MPa ( 4 4 d d ) ( 4 4 π ) π The torque Τ produces shear stresses at points A and B. d T ( 7.kN. m)( 0.m) π ( d d ) π ( ) Torsion 4 MPa Finally, we need to calculate the direct shear stresses at points A and B due to the shear force V.
16 The shear stress at point A is zero, and the shear stress at point B ( ) is obtained from the shear formula for a circular tube ( 4800) V, Ma MPa A 0.057m The stresses acting on the cross section at points A and B have now been calculated. VQ Ib π I Q b ( r 4 4 ( d d ) 64 ( r r ) r )
17 Stress Elements For both elements the y-ais is parallel to the longitudinal ais of the pole and the -ais is horizontal. Point A : 0 y a 54.9MPa y 6.4MPa Principal stresses at Point A Substituting, 7.5MPa +/- 8.MPa 55.7MPa and - 0.7MPa + y y, + ± ( ) y The maimum in-plane shear stresses can be obtained from the equation y MAX + y ( ) 8. MPa Because the principal stresses have opposite signs, the maimum in-plane shear stresses are larger than the maimum out-of-plane shear stresses. Then, ma 8.MPa.
18 Point B : y 0 y + y 6.4MPa MPa 7.0MPa Principal stresses at point B are 7.0MPa MPa And the maimum in-plane shear stress is ma 7.0MPa The maimum out-of-plane shear stresses are half of this value. Note If the largest stresses anywhere in the pole are needed, then we must also determine the stresses at the critical point diametrically opposite point A, because at that point the compressive stress due to bending has its largest value. The principal stresses at that point are 0.7MPa and MPa The maimum shear stress is 8.MPa. (In this analysis only the effects of wind pressure are considered. Other loads, such as weight of the structure, also produce stresses at the base of the pole).
19 A tubular post of square cross section supports a horizontal platform (see figure). The tube has outer dimension b 6in. And wall thickness t 0.5in. The platform has dimensions 6.75in 4.0in and supports an uniformly distributed load of 0psi acting over its upper surface. The resultant of this distributed load is a vertical force P (0psi)(6.75in 4.0in) 40lb This force acts at the midpoint of the platform, which is at distance d 9in. from the longitudinal ais of the post. A second load P 800lb acts horizontally on the post at height h 5in above the base. Determine the principal stresses and maimum shear stresses at points A and B at the base of the post due to the loads P and P.
20 Solution Stress Resultants The force P acting on the platform is statically equivalent to a force P and a moment M P d acting on the centroid of the cross section of the post. The load P is also shown. The stress resultant at the base of the post due to the loads P and P and the moment M are as follows: (A) An aial compressive force P 40lb (B) A bending moment M produced by the force P : M P d (40lb)(9in) 960lb-in (C) A shear force P 800lb (D) A bending moment M produced by the force P : M P h (800lb)(5in) 4600lb.in Eaminations of these stress resultants shows that both M and M produce maimum compressive stresses at point A and the shear force produces maimum shear stresses at point B. Therefore, A and B are the critical points where the stresses should be determined.
21 Stresses at points A and B (A) The aial force P produces uniform compressive stresses throughout the post. These stresses are P P / A where A is the cross section area of the post A b (b t) 4t(b-t) 4 (0.5in)(6in 0.5in).0in P P / A 40lb /.00in 95psi (B) The bending moment M produces compressive stresses M at points A and B. These stresses are obtained from the fleure formula M M (b / ) / Ι where Ι is the moment of inertia of the cross section. The moment of inertia is Ι [b 4 -(b -t) 4 ] / [(6in) 4 (5in) 4 ] / 55.9in 4 Thus, M M b / Ι (960lb.in)(6in) / ()(55.9in 4 ) 564psi
22 (C) The shear force P produces a shear stress at point B but not at point A. We know that an approimate value of the shear stress can be obtained by dividing the shear force by the web area. P P / A web P /(t(b t)) 800lb / ()(0.5in)(6in in) 60psi The stress p acts at point B in the direction shown in the above figure. We can calculate the shear stress P from the more accurate formula. The result of this calculation is P 6psi, which shows that the shear stress obtained from the approimate formula is satisfactory. D) The bending moment M produces a compressive stress at point A but no stress at point B. The stress at A is M M b / Ι (4600lb.in)(6in) / ()(55.9in 4 ) psi. This stress is also shown in the above figure.
23 Stress Elements Each element is oriented so that the y-ais is vertical (i.e. parallel to the longitudinal ais of the post) and the -ais is horizontal ais Point A : The only stress in point A is a compressive stress a in the y direction A P + M + M A 95psi + 564psi + psi 4090psi (compression) Thus, this element is in uniaial stress. Principal Stresses and Maimum Shear Stress 0 y - a psi y 0 Since the element is in uniaial stress, and y psi And the maimum in-plane shear stress is ma ( - ) / ½ (4090psi) 050psi The maimum out-of-plane shear stress has the same magnitude.
24 Point B: Here the compressive stress in the y direction is B P + M B 95psi + 564psi 860psi (compression) And the shear stress is B P 60psi The shear stress acts leftward on the top face and downward on the face of the element. Principal Stresses and Maimum Shear Stress 0 + y y y - B - 860psi, ± + y - P - 60psi Substituting, - 90psi +/- 944psi 4psi and - 870psi The maimum in-plane shear stresses can be obtained from the equation Because the principal stresses have opposite signs, the maimum in-plane shear stresses are larger than the maimum out-of-plane shear stresses. Then, ma 944psi. + ( ) psi MAX y 944 ( ) y
25 Three forces are applied to a short steel post as shown. Determine the principle stresses, principal planes and maimum shearing stress at point H. Solution Determine internal forces in Section EFG. V M M M y 0 kn ( 50kN)( 0.0 m) ( 75kN)( 0.00 m) 8.5kN m 0 M z P 50kN V z 75kN ( 0kN)( 0.00 m) kn m
26 Note: Section properties, Evaluate the stresses at H. Shear stress at H. Q A y yz V I z Q t [( m)( 0.045m) ]( m) MPa m ( )( 6 ) 75kN m ( m )( m) A I I z y ( m)( 0.40 m) + ( m)( 0.40 m) m ( 0.40 m)( m) m Normal stress at H. P A 50kN m + M I z z a + ( 8.5kN m)( 0.05m) M I m ( kn m)( 0.00 m) m m ( ) MPa 66.0MPa b
27 Calculate principal stresses and maimum shearing stress. ma ma min tan θ p.98 θ p p ma R ma min OC OC + CY CD θ R R MPa 70.4 MPa 7.4 MPa p 7.4 MPa 70.4 MPa 7.4 MPa 7.96
28 The cantilever tube shown is to be made of 04 aluminum alloy treated to obtain a specified minimum yield strength of 76MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n4. The bending load is F.75kN, the aial tension is P9.0kN and the torsion is T7N.m. What is the realized factor of safety? Consider the critical area ( top surface).
29 S y 0.76 GPa 0. GPa VM n P A + Mc I Maimum bending moment 0F 9kN A z VM Tr J 0mm.75kN + I d 7 J + 6d J ( ) z For the dimensions of that tube d n S y VM
30 A certain force F is applied at D near the end of the 5-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 05 steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 8kpsi. Find the force (F) required to initiate yielding. Assume that the lever DC will not yield and that there is no stress concentration at A. Solution: ) Find the critical section The critical sections will be either point A or Point O. As the moment of inertia varies with r 4 then point A in the in diameter is the weakest section.
31 ) Determine the stresses at the critical section ) Chose the failure criteria. The AISI 05 is a ductile material. Hence, we need to employ the distortion-energy theory. + F VM S y VM y My I d M πd 64 F 4in πd Tr J d T πd 6 F 5in π (in) z 4 y 46lbf + y + z 94.5F F F
32 Apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two nonzero principal stresses ( A and B ) will have opposite signs (Case ). ( ) ( ) ( ) lbf F F F S S z z y B A z y B A ma ±
33 A round cantilever bar is subjected to torsion plus a transverse load at the free end. The bar is made of a ductile material having a yield strength of 50000psi. The transverse force (P) is 500lb and the torque is 000lb-in applied to the free end. The bar is 5in long (L) and a safety factor of is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid yielding using both MSS and DET criteria. Solution ) Determine the critical section The critical section occurs at the wall.
34 4 64 d PL d d PL I Mc π π 4 6 d T d d T J Tc y π π ( ) ( ) ( ) ( ) + ± + ± + ± + ± + ± +,,, d T PL PL d d T d PL d PL y y y y π π π π π
35 6450 d 6450 d MAX d d MSS d.0 in MAX The stresses are in the wrong order.. Rearranged to 6450 S n y ( 980.8) 75 d d ,000.4 DET VM VM d d.05in S n y d d 6450 d d
36 The factor of safety for a machine element depends on the particular point selected for the analysis. Based upon the DET theory, determine the safety factor for points A and B. This bar is made of AISI 006 cold-drawn steel (S y 80MPa) and it is loaded by the forces F0.55kN, P8.0kN and T0N.m Solution: d Fl 4 Point A Mc P P Fl P I Area πd πd πd + πd 64 4 ( )( )( 0.) π ( 0.0) ( )( 8 0 ) MPa π ( 0.0)
37 n VM S y VM Tr J 6T πd 6( 0) ( 0.00) y 9. 0 π [ ] ( ) ( 9.) y.77 Point B ( )( 4P ) n y VM πd 6T πd ( 0.0) 6( 0) A π ( 0.0) [ 5.47 (.4) ] V + π MPa ( )( ) 4 + π 4 ( 0.0) MPa 0.MPa 45.0MPa.4MPa
38 The shaft shown in the figure below is supported by two bearings and carries two V- belt sheaves. The tensions in the belts eert horizontal forces on the shaft, tending to bend it in the -z plane. Sheaves B eerts a clockwise torque on the shaft when viewed towards the origin of the coordinate system along the -ais. Sheaves C eerts an equal but opposite torque on the shaft. For the loading conditions shown, determine the principal stresses and the safety factor on the element K, located on the surface of the shaft (on the positive z-side), just to the right of sheave B. Consider that the shaft is made of a steel of a yield strength of 8ksi
40 Shearing force 65lb Bending Moment -540lb-in Mc M ( r) I πr π ( 0.65) 0ksi 4 ( ) ( 0.65) Tr T 00 z. 868ksi J πr π
41 ( ) ( ) ( ) ksi ksi y y y y ,, + ± + ± + ± + ( ) Ma Y Ma S n Factor Safety ksi MSS ( ) ( ) ( )( ) VM y VM S n Factor Safety ksi DET
42 A horizontal bracket ABC consists of two perpendicular arms AB and BC, of.m and 0.4m in length respectively. The Arm AB has a solid circular cross section with diameter equal to 60mm. At point C a load P.0kN acts vertically and a load P.07kN acts horizontally and parallel to arm AB. For the points p and q, located at support A, calculate: ()The principal stresses. () the maimum in-plane shear stress..m q
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05 Solutions 46060 5/5/10 3:53 PM Page 14 010 Pearson Education, Inc., Uer Saddle River, N. ll rights reserved. his material is rotected under all coyright laws as they currently exist. No ortion of this
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Homework 9 Problems: 1.31, 1.3, 14.4, 14.1 Problem 1.31 Assume that if the shear stress exceeds about 4 10 N/m steel ruptures. Determine the shearing force necessary (a) to shear a steel bolt 1.00 cm in