Stresses and Maximum Shear Stresses at the Selected Point

Home , Bending moment

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. Examples: wide‐flange beam supported by a cable (combined bending and axial load), cylindrical pressure vessel supported as a beam, and shaft in combined torsion and bending. Method of Analysis: 1.Select the point on the structure where the stresses and the strains are to be determined. 2.For each load on the structure, determine the stress resultant at the cross section containing the selected point.. 3.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 maximum 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 Τρ My σ = τ = σ −= A Ιρ I VQ pr τ = σ = Ib t 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. The twisting moment Τ produces a torsional shear stresses Tr 2T τ torsion == 3 I Polar πr

The stress τ1 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 4M σ == bending I πr3 However, the bending moment produces no stress at point B, because B is located on the neutral axis. 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: VQ 4V τ == shear Ib 3A

σA and τ1 are acting in point A, while the τ1 and τ2 are acting in point B. Note that the element is in plane stress with

σx = σA, σy = 0 , and τxy = - τ1. A stress element in point B is also in plane stress and the only stresses acting on this element are the shear

stresses τ1 and τ2. Therefore

σx = σy = 0 and τxy = - ( τ1 + τ2).

At point A: σx = σA, σy = 0 , and τxy = - τ1

At point B σx = σy = 0 and τxy = - ( τ1 + τ2). Of interest are the points where the stresses calculated from the flexure and shear formulas have maximum or minimum values, called critical points.

For instance, the normal stresses due to bending are largest at the cross section of maximum bending moment, which is at the support. Therefore, points C and D at the top and bottom of the beam at the fixed 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. Stress at which point? 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 axial loading. For a 50mm diameter shaft transmitting a torque Τ = 2.4kN.m and a tensile force P = 125kN, determine the maximum tensile stress, maximum compressive stress, and maximum shear stress in the shaft.

Solution The stresses in the rotor shaft are produced by the combined action of the axial force P and the torque Τ. Therefore the stresses at any point on the surface of the shaft consist of a tensile stress σo and a shear stress τo. P 125kN The tensile σ == = 66.63 MPa stress A π ()05.0 m 2 4 ⎛ 05.0 ⎞ ().4.2 mkN ⎜ ⎟ Tr 2 The shear stress to is obtained τ == ⎝ ⎠ = 78.97 MPa from the torsion formula Torsion 4 I P π ()05.0 32 Knowing the stresses σo and τo, we can now obtain the principal stresses and maximum shear stresses . The principal stresses are obtained from 2 ⎛ + ⎞ ⎛ −σσσσ⎞ ⎜ yx ⎟ ⎜ yx ⎟ ()2 σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ + τ xy ⎝ 2 ⎠ ⎝ 2 ⎠

2 ⎛ + 66.630 ⎞ ⎛ − 66.630 ⎞ 2 σ = ⎜ ⎟ ± ⎜ ⎟ ()−+ 78.97 2,1 2 2 The maximum in-plane shear ⎝ ⎠ ⎝2 ⎠ ⎛ −σσ⎞ ⎜ yx ⎟ ()2 τ MAX = ⎜ ⎟ τ xy =+ 103MPa stresses are obtained using the σ1 = 135MPa ⎝ 2 ⎠ formula σ 2 −= 71MPa

Because the principal stresses σ1 and σ2 have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses. Therefore, the maximum shear stress in the shaft is 103MPa.

Will it fail if σyield=480MPa?

480MPa σ ()()()()2 2 =−+−−= 2.1817171135135 MPa MSST SF =⇒ 2 = 33.2 VM 103MPa 480MPa SFDET =⇒ = 65.2 2.181 MPa A thin wall cylindrical pressure vessel with a circular cross section is subjected to internal gas pressure p and simultaneously compressed by an axial load P = 12k. The cylinder has inner radius r = 2.1in. And wall thickness t

= 0.15in. Determine the maximum allowable internal pressure pallow 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 axial force. We can isolate a stress element in point A. The x-axis is parallel to the longitudinal axis of the pressure vessel and the y-axis is circumferential. Note that there are no shear stresses acting on the element.

The longitudinal stress σx is equal to the tensile stress produced by the internal pressure minus the compressive stress produced by the axial force. pr P pr P σ −=−= x t A t 222 πrt The circumferential stress is equal to the tensile pr σy σ = stress produced by the internal pressure. y t Note that σ > σ . y x pr Since no shear stresses act on the element the 1 σσy == t above stresses are also the principal stresses pr P 2 σσx −== substituting numerical values t 22 πrt pr ()12 in.p σ == = 0.14 p 1 t 150 in. pr P ()12 in.p 12k σ =−= − −= 60630.7 psip 2 t 22 πrt ()1502 in. []π ()()15.01.22 inin

In-Plane Shear Stresses The maximum in-plane shear stress is ()σ −σ ()()− ( − 60630.70.14 psipp ) τ = 21 = += 30325.3 psip Max 2 2

Since τmax is limited to 6500psi then

65000 = + 30324.3 ⇒ ppsippsi allowed = 990 psi Out-of-Plane Shear Stresses σ τ = 1 The maximum out-of-plane shear stress is either Max 2 From the first equation we get σ τ = 2 65000 = − 30325.3 psippsi Max 2

pallowed = 2720 psi From the second equation we 65000 = 0.7 ⇒ pppsi allowed = 928psi get:

Allowable internal pressure Comparing the three calculated values for the allowable pressure, we see that (pallow)3 = 928psi governs. At this pressure, the principal stresses are

σ1 = 13000psi and

σ2 = 430psi. These stresses have the same signs, thus confirming that one of the out- of-plane shear stresses must be the largest shear stress. A sign of dimensions 2.0mx1.2m is supported by a hollow circular pole having outer diameter 220mm and inner diameter 180mm (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 maximum shear stresses at points A and B at the base of the pole due to wind pressure of 2.0kPa 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: = pAW = ( )( × )= 8.42.10.20.2 kNmmkPa

The line of action of this force is at height h = 6.6m above the ground and at distance b = 1.5m 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. The torque is equal to the force W times the distance b: = WbT = ( )( 5.18.4 mkN ) T = 2.7 N − mk

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) = 31.68kN.m Τ = 7.2kN.m V = W = 4.8kN Examination of these stress resultants shows that maximum bending stresses occur at point A and maximum 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 σa at point A, but no stress at point B (which is located on the neutral axis) ⎛ d ⎞ M ⎜ 2 ⎟ 2 ()()11.068.31 mkN ⎝ ⎠ 91.54 MPa σ a = 4 4 = 4 4 = ⎡π ()2 − dd 1 ⎤ ⎡π ()− 18.022.0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ 64 ⎦ ⎣ 64 ⎦

The torque Τ produces shear stresses τ1 at points A and B. ⎛ d ⎞ T⎜ 2 ⎟ 2 ()()11.0.2.7 mmkN ⎝ ⎠ 24.6 MPa τTorsion = 4 4 = 4 4 = ⎡π ()2 − dd 1 ⎤ ⎡π ()− 18.022.0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ 32 ⎦ ⎣ 32 ⎦ Finally, we need to calculate the direct shear stresses at points A and B due to the shear force V. VQ The shear stress at point A is zero, and the shear stress at τ = 2 Ib point B (τ ) is obtained from the shear formula for a circular 2 4 4 tube ⎡π ()2 − dd 1 ⎤ V (480022 ) I = ⎢ ⎥ τ == = 7637.0 MPa ⎣ 64 ⎦ ,2 Max A 01257.0 m2 2 3 3 ()2 −= rrQ 1 The stresses acting on the cross section at points A and B 3 have now been calculated. −= rrb 12 )(2 Stress Elements For both elements the y-axis is parallel to the longitudinal axis of the pole and the x-axis is horizontal. Point A :

σx = 0

σy = σa = 54.91MPa τxy = τ1 = 6.24MPa

2 ⎛ + ⎞ ⎛ −σσσσ⎞ Principal stresses at Point A ⎜ yx ⎟ ⎜ yx ⎟ ()2 σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ + τ xy ⎝ 2 ⎠ ⎝ 2 ⎠

Substituting σ1,2 = 27.5MPa +/- 28.2MPa

σ1 = 55.7MPa and σ2 = - 0.7MPa 2 The maximum in-plane shear ⎛ − σσ⎞ ⎜ yx ⎟ 2 stresses can be obtained from τ MAX = ⎜ ⎟ ()τ xy =+ 2.28 MPa ⎝ 2 ⎠ the equation Because the principal stresses have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses.

Then, τmax = 28.2MPa. Point B :

σx = σy = 0

τxy = τ1 + τ2

τxy = 6.24MPa + 0.76MPa = 7.0MPa Principal stresses at point B are

σ1 = 7.0MPa σ2 = - 7.0 MPa And the maximum in-plane shear stress is

τmax = 7.0MPa The maximum 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

σ1 = 0.7MPa and σ2 = - 55.7MPa The maximum shear stress is 28.2MPa. (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). 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 x 24.0in and supports an uniformly distributed load of 20psi acting over its upper surface. The resultant of this distributed load is a vertical force

P1 = (20psi)(6.75in x 24.0in) = 3240lb This force acts at the midpoint of the platform, which is at distance d = 9in. from the longitudinal axis of the post.

A second load P2 = 800lb acts horizontally on the post at height h = 52in above the base. Determine the principal stresses and maximum shear stresses at points A and B at the base of the post due to the loads P1 and P2. Solution Stress Resultants

The force P1 acting on the platform is statically equivalent to a force P1 and a moment

M1 = P1d acting on the centroid of the cross section of the post.

The load P2 is also shown. The stress resultant at the base of the post due to the

loads P1 and P2 and the moment M1 are as follows:

(A) An axial compressive force P1 = 3240lb

(B) A bending moment M1 produced by the force P1:

M1 = P1d = (3240lb)(9in) = 29160lb-in

(D) A bending moment M2 produced by the force P2:

M2 = P2h = (800lb)(52in) = 41600lb.in

Examinations of these stress resultants shows that both M1 and M2 produce maximum compressive stresses at point A and the shear force produces maximum shear stresses at point B. Therefore, A and B are the critical points where the stresses should be determined. Stresses at points A and B

(A) The axial force P1 produces uniform compressive stresses throughout the post. These stresses are σP1 = P1 / A where A is the cross section area of the post A = b2 –(b –2t)2 = 4t(b-t) = 4 (0.5in)(6in – 0.5in) = 11.0in2 2 σP1 = P1 / A = 3240lb / 11.00in = 295psi

(B) The bending moment M1 produces compressive stresses σM1 at points A and B. These stresses are obtained from the flexure formula

σM1 = M1 (b / 2) / Ι where Ι is the moment of inertia of the cross section. The moment of inertia is Ι = [b4 -(b -2t)4] / 12 = [(6in)4 – (5in)4] / 12 = 55.92in4 4 Thus, σM1 = M1 b / 2Ι = (29160lb.in)(6in) / (2)(55.92in ) = 1564psi (C) The shear force P2 produces a shear stress at point B but not at point A. We know that an approximate value of the shear stress can be obtained by dividing the shear force by the web area.

τP2 = P2 / Aweb =P2 /(2t(b – 2t)) =800lb / (2)(0.5in)(6in–1in)= 160psi

The stress τp2 acts at point B in the direction shown in the above figure.

We can calculate the shear stress τP2 from the more accurate formula. The result of this calculation is τP2 = 163psi, which shows that the shear stress obtained from the approximate formula is satisfactory.

D) The bending moment M2 produces a compressive stress at point A but no stress at point B. The stress at A is 4 σM2 = M2 b / 2Ι = (41600lb.in)(6in) / (2)(55.92in ) = 2232psi. This stress is also shown in the above figure. Stress Elements Each element is oriented so that the y-axis is vertical (i.e. parallel to the longitudinal axis of the post) and the x-axis is horizontal axis Point A : The only stress in point A is a

compressive stress σa in the y direction

σA = σP1 + σM1 + σM2

σA = 295psi + 1564psi + 2232psi = 4090psi (compression) Thus, this element is in uniaxial stress. Principal Stresses and Maximum Shear Stress

σx = 0

σy = - σa = - 4090psi

τxy = 0 Since the element is in uniaxial stress,

σ1 = σx and σ2 = σy = - 4090psi And the maximum in-plane shear stress is

τmax = (σ1 - σ2) / 2 = ½ (4090psi) = 2050psi The maximum out-of-plane shear stress has the same magnitude. Point B: Here the compressive stress in the y direction is

σB = σP1 + σM1 σB = 295psi + 1564psi = 1860psi (compression) And the shear stress is

τB = τP2 = 160psi The shear stress acts leftward on the top face and downward on the x face of the element. Principal Stresses and Maximum Shear Stress

σx =0 2 ⎛ + ⎞ ⎛ −σσσσ⎞ σ = - σ = - 1860psi ⎜ yx ⎟ ⎜ yx ⎟ ()2 y B σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ + τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ τxy = - τP2 = - 160psi

Substituting σ1,2 = - 930psi +/- 944psi

σ1 = 14psi and σ2 = - 1870psi

The maximum in-plane shear stresses can be obtained from the equation Because the principal stresses have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses. σ 2 Then, τmax = 944psi. ⎛ x ⎞ 2 τ MAX = ⎜ ⎟ ()τ xy =+ 944 psi ⎝ 2 ⎠ Three forces are applied to a short steel post as shown. Determine the principle stresses, principal planes and maximum shearing stress at point H. Solution Determine internal forces in Section EFG.

Vx −= 30 P = kN50kN Vz = − kN75

M x = ()()()()− m200.0kN75m130.0kN50

M x ⋅−= mkN5.8

M y = 0 M z = ()()⋅= mkN3m100.0kN30 A = ( )( ) ×= − m106.5m140.0m040.0 23 Note: Section properties, I = 1 ()()3 ×= − m1015.9m140.0m040.0 46 x 12 Evaluate the stresses at H. I = 1 ()()3 ×= − m10747.0m040.0m140.0 46 z 12 Normal stress at H.

P z aM x bM σ y −++= A Iz Ix kN50 ()()⋅ m020.0mkN3 = + × m105.6 23- × − m10747.0 46 ()()⋅ m025.0mkN5.8 Shear stress at H. − × − m1015.9 46 yAQ 11 == []()()( m0475.0m045.0m040.0 ) ()−+= = MPa66.0MPa2.233.8093.8 ×= − m105.85 36

QV ()()× − m105.85kN75 36 τ z == yz − 46 xtI ()× ()m040.0m1015.9 = MPa52.17 Calculate principal stresses and maximum shearing stress.

22 τmax R =+== MPa4.3752.170.33

σ max ROC =+=+= MPa4.704.370.33

σ min ROC −=−=−= MPa4.74.370.33 CY 52.17 2tan θ == θ 96.272 °= p CD 0.33 p

θ p 98.13 °=

τmax = MPa4.37

σ max = MPa4.70

σ min −= MPa4.7

θ p 98.13 °= The cantilever tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n=4. The bending load is F=1.75kN, the axial tension is P=9.0kN and the torsion is T=72N.m. What is the realized factor of safety?

Consider the critical area ( top surface). S 276.0 σ y =≤ GPa = 0690.0 GPa VM n 4 P Mc σ += x A I Maximum bending moment = 120F ⎛ d ⎞ mm× 75.1120 kNx⎜ ⎟ 9kN 2 σ += ⎝ ⎠ x A I ⎛ d ⎞ 72×⎜ ⎟ Tr 2 36d τ == ⎝ ⎠ = zx J J J

1 22 2 VM ()+= 3τσσzxx For the dimensions of that tube S 0.276 n = y = = 4.57 σVM 0.06043 A certain force F is applied at D near the end of the 15-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 1035 steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 81kpsi. 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: 1) Find the critical section The critical sections will be either point A or Point O. As the moment of inertia varies with r4 then point A in the 1in diameter is the weakest section. 2) Determine the stresses at the ⎛ d ⎞ critical section M ⎜ ⎟ My 2 ×× 1432 inF σ == ⎝ ⎠ = = 6.142 F x I πd 4 πd 3 64 3) Chose the failure ⎛ d ⎞ criteria. T⎜ ⎟ Tr 2 ×× 1516 inF τ == ⎝ ⎠ = = 4.76 F The AISI 1035 is a zx J πd 4 π in)1( 3 ductile material. Hence, we need to employ the 32 distortion-energy theory. 22 2 2 2 VM 3 xyyxyx x τστσσσσσzx =+=+−+= 5.1943 F S 81000 F y === 416lbf σVM 5.194 Apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two non- zero principal stresses (σA and σB) will have opposite signs (Case 2).

2 −σσBA S y ⎛σ x ⎞ 2 τ max = ±== ⎜ ⎟ +τ zx ⎝ 222 ⎠

2 ⎛σ x ⎞ 2 2 2 σσS yBA =≥− 2 ⎜ ⎟ zx x +=+ 4τστzx ⎝ 2 ⎠ 1 81000 = ()()()F 2 ×+ 4.7646.142 F 2 2 = 388lbfF 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 1000lb-in applied to the free end. The bar is 5in long (L) and a safety factor of 2 is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid yielding using both MSS and DET criteria.

Solution 1) Determine the critical section

The critical section occurs at the wall. ⎛ d ⎞ ⎛ d ⎞ PL⎜ ⎟ T⎜ ⎟ Mc ⎝ 2 ⎠ 32PL Tc ⎝ 2 ⎠ 16T σ x == = τ == = I πd 4 πd 3 xy J πd 4 πd 3 64 32

2 2 ⎛ + yx ⎞ ⎛ −σσσσyx ⎞ 2 ⎛ ⎞ ⎛σσ⎞ 2 ⎜ ⎟ ⎜ ⎟ x x σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ ()τ xy =+ ⎜ ⎟ ± ⎜ ⎟ + ()τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎝ 22 ⎠

2 2 16PL ⎛16PL ⎞ ⎛ T ⎞ 1616 ⎡ 2 2 ⎤ σ 2,1 ±= ⎜ ⎟ + ⎜ ⎟ ()+±= TPLPL d 3 ⎝ d 3 ⎠ ⎝ 3 ⎠ ππππdd 3 ⎣⎢ ⎦⎥ 16 σ = ⎡ ()2 +×±× 100055005500 2 ⎤ 2,1 πd 3 ⎣⎢ ⎦⎥ 26450 8.980 The stresses are in the wrong σ = σ −= 1 d 3 2 d 3 order.. Rearranged to 26450 8.980 σ = σ −= 1 d 3 3 d 3 σ − σ − − )8.980(26450 4.13715 τ = 31 = = MSS MAX 2 2d 3 d 3 S 50000 2τσσ y ==≤=− 000,25 31 MAX n 2 d ≥ 031.1 in

2 2 2 2 ⎛ 26450 ⎞ ⎛ 8.980 ⎞ ⎛ 26450 ⎞⎛ 8.980 ⎞ σσσσσ=−+= ⎜ ⎟ ⎜−+ ⎟ − ⎜ ⎟⎜− ⎟ VM 1 3 31 d 3 d 3 3 dd 3 DET ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ 26950 S 50000 σ y =≤= VM d 3 n 2 ≥ 025.1 ind 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 1006 cold-drawn steel (Sy=280MPa) and it is loaded by the forces F=0.55kN, P=8.0kN and T=30N.m

⎛ d ⎞ Solution: Fl⎜ ⎟ Mc P 2 P Fl 432 P Point A σ =+= ⎝ ⎠ +=+ x I Area d 4 ππd 2 d 3 ππd 2 464

()( 3 )( 1.01055.032 ) ( )(1084 3 ) σ = + = 49.95 MPa x π ()02.0 3 π ()02.0 2 Tr T (301616 ) τ === = 10.19 MPa xy J πd 3 π ()020.0 3 1 22 2 2 2 VM ()τσσxyx [ +=+= ()] = 1.1011.19349.953 MPa S 280 n y === 77.2 σVM 1.101

Point B P ( )(10844 3 ) σ == = 47.25 MPa x πd 2 π ()02.0 2 T 416 V ()3016 ()()1055.04 3 τ xy 3 =+= 3 + = 43.21 MPa πd 3A π ()02.0 ⎛ π ⎞ 2 3⎜ ⎟()02.0 ⎝ 4 ⎠ 1 2 2 2 σVM []+= ()= 02.4543.21347.25 MPa 280 n == 22.6 02.45 The shaft shown in the figure below is supported by two bearings and carries two V- belt sheaves. The tensions in the belts exert horizontal forces on the shaft, tending to bend it in the x-z plane. Sheaves B exerts a clockwise torque on the shaft when viewed towards the origin of the coordinate system along the x-axis. Sheaves C exerts 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 81ksi

Shearing force = 165lb Bending Moment = -1540lb-in Mc (rM ) −1540 σ −=−=−= 4 = 031.8 ksi x I πr 4 π ()625.0 3 4

Tr T (110022 ) τ === = 868.2 ksi xz J πr3 π ()625.0 3 2 2 ⎛ + yx ⎞ ⎛ −σσσσyx ⎞ 2 ⎛ x ⎞ ⎛σσx ⎞ 2 σ 2,1 = ⎜ ⎟ ± ⎜ ⎟ ()τ xy =+ ⎜ ⎟ ± ⎜ ⎟ + ()τ xy ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎝ 22 ⎠

2 03.8 ⎛ 03.8 ⎞ 2 σ 2,1 ±= ⎜ ⎟ + ()868.2 2 ⎝ 2 ⎠

σ1 = 95.8 ksi σ 2 −= 92.0 ksi

σ −σ − (− 92.095.8 ) MSS.....τ = 31 = = 935.4 ksi Max 2 2

SY 81 Safety.. nFactor 2 2 ==== 2.8 τ Max 935.4

2 2 2 2 DET...... VM 1 3 σσσσσ31 ()( )()()=−−−+=−+= 44.992.095.892.095.8 ksi S 81 Safety.. nFactor y ==== 58.8 σVM 44.9 A horizontal bracket ABC consists of two perpendicular arms AB and BC, of 1.2m 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 P1=2.02kN acts vertically and a load P2=3.07kN acts horizontally and parallel to arm AB. For the points p and q, located at support A, calculate: (1)The principal stresses. (2) the maximum in-plane shear stress.

1.2m