Two forces PI and Pz, of magnitude PI::::: 15 kN andP2 ::::: 18 leN, are applied as shown to the end A of bar AB,which is welded to a cylindrical member BD of radiusc ::::: 20 mm (Fig. 8.21). Knowing that the distance from A tothe axis of member BD is a ::::: 50 mm and assuming that allslresses remain below the propOrtional ]imit of the material,determine (a) the normal and shearing stresses at point K ofthe transverse section of member BD located at a distanceb ::::: 60 mm from end B, (b) the principal axes and principalstresses at K, (c) the maximum shearing stress at K.

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SAMPLE PROBLEM 8.4

A horizontal 500-lb force acts at point D of crankshaft AB which is held instatic equilibrium by a twisting couple T and by reactions at A and B. Know-ing that the bearings are self-aligning and exer~ no couples on the shaft, de-termine the normal and shearing stresses at points H, J, K, and L located at theends of the vertical and horizontal diameters of a transverse section located2.5 in. to the left of bearing B.

SAMPLE PROBLEM 8.4

SOLUTION

Free Body. Entire Crankshaft. A = B = 250 Ib

+ ~ 2:M~ = 0: -(500 Ib)(1.8 in.) + T = 0 T = 9001b . in.

Internal Forces in Transverse Section. We replace the reaction B andthe twisting couple T by an equivalent force-couple system at the center C ofthe transverse section containjng H. J, K, and L.

%

v = B = 250 Ib T = 900 Ib . in.My = (250 Ib)(2.5 in.) = 6251b . in.

The geometric properties of the O.9-in.-diameter section are

A = 7T(O.45 in.)2 = 0.636 in2 1= !7T(O.45 in.)4 = 32.2 X 10-3 in4J = !7T(O.45 in.)4 = 64.4 X 10-3 in4

Stresses Produced by Twisting Couple T. Using Eq. (3.8), wedetermine the shearing stresses at points H, J, K, and L and show them inFig. (a).

Tc (900 Ib . in.)(0.45 in.) = 6290 psi'T = J = 64.4 X 10 3 in4

Stresses Produced by Shearing Force V. The shearing force V pro-duces no shearing stresses at points J and L. At points Hand K we first com-pute Q for a semicircle about a vertical diameter and then determine the shear-ing stress produced by the shear force V = 250 lb. These stresses are shownin Fig. (b).

Q = (.!.1Tc2)( 4C)= ~c3 = ~(O 45 in)3 = 60 7 X 10-3 in3

,2 31T 33' . .

VQ (250 Ib)(60.7 X 1O-3in3) .T = - ::::: = 524 pSi

It (32.2 X 10-3 in4)(O.9 in.),

Stresses Produced by the Bending Couple M,.. Since the bending cou-ple My acts in a horizontal plane, it produces no stresses at Hand K. Using Eq.(4.15), we detennine the normal stresses at points J and L and show them in

Fig. (c).

IMylc (625lb . in.)(O.45 in.) .(J' = - = = 8730PSi

I 32.2 X 10-3 in4

Summary. We add the stresses shown and obtain the total normal andshearing stresses at points H. J, K, and L

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SAMPLE PROBLEM 8.5.

Three forces are applied as shown at points A, B, and Jj of a short steel post.Knowing that the horizontal cross section of the post is a 40 X 140-mm rec-tangle, determine the principal stresses, principal planes and maximum shear-ing stress at point H. .

~

SOLUTION

Internal Forces in Section EFG. We replace the three applied forcesby an equivalent force-couple system at the center C of the rectangular sectionEFG. We have. .

Vx = -30 kN P = 50kN V~ = -75 kNMx = (50 kN)(0.130 m) - (75 kN)(0.200 m) = -8.5 kN . mMy = 0 M~ = (30 kN)(0.1O0 m) = 3 kN . m

We note that there is no twisting couple about the y axis. The geometricproperties of the rectangular section are

A= (0.040 m)(0.140 m) = 5.6 X 10-3 m2Ix= i2(0.040 m)(0.140 m? = 9.15 X 10-6 m4

lz = i2(0.140 m)(0.040 m? = 0.747 X 10-6 m4

Normal Stress at H. We note that normal stresses (J'y are produced bythe centric force P and by the bending couples Mx and M<;. We determine thesign of each stress by carefully examining the sketch of the force-couple sys-tem at C.

P IMtla IMx1b(J' = +-+---

Y A I I,t x

50 kN (3 kN . m)(0.020 m) (8.5 kN . m)(O.O25 m)= + -

5.6 X 10-3 m2 0.747 X 1O-6m4 9.15 X 10-6 m4Uy = 8.93 MPa + 80.3 MPa - 23.2 MPa cry = 66.0 MPa. <1

Shearing Stress at H. Considering first the shearing force Vx, we notethat Q = 0 with respect to the z axis,. since H is on the edge of the cross sec-tion. Thus V x produces no shearing stress at H. The shearing force Vz does pro-duce a shearing stress at H and we write

Q = A1YJ ~ [(0.040 m)(0.045 m)](0.0475 m) = 85.5 X 10-6 m3

VzQ (75 kN)(85.5 X 10-6 m3) .Tyz= Ixl= (9.15 X 1O-6m4)(0.040m) Tyz= 17.52MPa <J

Principal Stresses~ Principal Planes, and Maximum Shearing Stressat H. We draw Mohr's circle for the stresses at point H