Period #11: STATICALLY INDETERMINATE SHAFTS

A. REVIEW

Tc T  Torsion Formula:  ; max JJ Tc d Maximum shear strain:  c max GJ dx T Incremental angle of twist: d  dx GJ

Fig. 11.1

B. STATICALLY INDETERMINATE SHAFTS Too many support conditions and not enough equations of static equilibrium. Or too many redundant members to carry the applied loads. Must generally augment conditions of static equilibrium with knowledge of how the indeterminate system deforms. This is called the kinematic constraint. Can use the principle of linear superposition to solve these systems.

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.1 TA

(a) (b) TB

Fig. 11.2 (a) Indeterminate shaft fixed at both ends; and (b) the free-body diagram.

From statics: TA+TB=T TTTAB Kinematic constraint: B/A=0 L BC 1 TT BA/  0 B LAC  CABC// T L TLTL TT =  AC  A AC B BC B L L JG JG 1 BC LAC TLTLA AC B BC L LBC BC T =T TTA  AB L LAC

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.2 Alternative Solution Using Principle of Superposition:

A A Case 2 Case 2: T TLTL  B AB B C  BA/ 2 Case 1 JG JG

TB Case 1: B TL B    AC BA/ 1 JG Combining the cases: 0   BABA//12  TL TL =AC + B JG JG

LAC L TTB  TT BC L A L

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.3 C. SHAFTS OF SOLID, NON-CIRCULAR CROSS-SECTIONS (TEXTBOOK, SECTION 5.6)

To this point, we’ve been concerned with shafts that have circular cross-sections.

What about shafts with non-circular cross-sections?

Consider the torsion of a rubber shaft with a square cross-section shown below (Fig. 11.3):

(a) undeformed square shaft (b) deformed square shaft when subjected to a torque

Figure 11.3 Twisting of rubber shaft with square cross-section

Note that the white square drawn on the square shaft “warps”. Some edges that were originally straight, become curved.

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.4 With solid non-circular cross-sections, the shear stress distribution over the cross-section cannot vary linearly with radial distance from the center (Fig. 11.4a). This leads to warping of the shaft cross-section (Fig. 11.4b).

(a) (b)

Fig. 11.4 Noncircular shaft response to torsion

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.5 Although the torsion formulae do not apply to shafts with non-circular cross-sections, the relevant properties of three sections are as indicated in the table below.

Table 11.1. Properties of non-circular sections.

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.6 D. THIN-WALLED CLOSED SECTION SHAFTS (TEXTBOOK 5.7)

Though you will not be responsible for this material, be aware that thin-walled, closed cross-sections can be treated using the concept of shear flow.

Fig. 11.5. Examples of thin-walled closed sections

T  qh ds qt constant = shear flow h = moment arm of segment ds about section centroid ds = infinitesimal length of wall segment.

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.7 , E. EXAMPLES Example 11.1: The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T=50lb.ft is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear and maximum shear strain in the brass and steel. Take

3 3 Gst 11.5(10 )ksi Gbr  5.6(10 )ksi

Along the length of the shaft the internal torque is T=50∙lbft.

In the composite section of the shaft, T=Tst + Tbr =-50∙lbft. 30.8Tbr T br   50 lb  ft

Tbr  1.57 lb  ft CBCB//    Kinematic constraint in the composite section CB: st br T 48.43 lb  ft TLTL st st BC br BC TL 600lbin 36 in  JG  JG BA BA    st br BA/    .001196rad  44 3  JG  4 6 2 1in .5 in 11.5 10 ksi AB in11.5 10 lb in   JG       2 TTTst 2  st br br T LBC 48.43  12 lb  in 24 in  JG  4 3 BC st    br .5in 5.6 10 ksi CB/  62     JG16.935 10 lb in 2  st 16,935k in2  .000824rad =Tbr2  30.8T br      .00202rad 550k in CABACB/// Tc 1.57 12 lb in .5in Tcst st 48.43 12 lb in 1in br br In BC segment    395psi In BC segment max     96 psi  max steel br  4 J  4 4 4 Jbr st 1 .5 in .5in 2   2 ENGR:2750 Mech. Def. Bodies The University of Iowa 11.8 Example 11.2. The two 3-ft-long shafts are made of 2014-T6 aluminum. Each has a diameter of 1.5in. and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts about their axes. If a torque of 600 lb-ft is applied to the top gear as shown, determine the maximum shear stress in each shaft. TA Statics: Kinematic constraint: T Gear F: f  B EF EEFFrr rF E f TA 36" TB FE ff EAEA    / 0  FE EF r JG F f EF TB T 36" Gear E:     0  B FBFB/ JG 600lb ft  TA  r E f FE

r EEFFrr =T  E T F ABr TT36"  36" F rrAB   EFJG JG 600lb ft =TAB 2 T rF 1 TTTABB rE 2 1 600lb ft  T  2 T 2 BB Tc 1.44k in .75 in Tc 2.88k in .75 in AE  BF   4.34ksi  max     2.17ksi  max BF T240 lb  ft  2.88 k  in AE  4 J  4 B J .75in .75in 2 2 TA 120 lb  ft  1.44 k  in

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.9 Example 11.3: The A-36 steel shaft is made from two segments: AC has a diameter of 0.5in and CB has a diameter T of 1 in. If the shaft is fixed at its ends A and B and subjected A to a uniform distributed torque of 60lb.in/in along segment CB, determine the absolute maximum shear stress in the shaft. Solve by method of linear superposition

Case 1: distributed torque-loading shown, free at support A Case 2: T  B to find A1. TA 3in 20 in A2  GJJst AC BC Case 2: Concentrated torque TA at A to find A2.  TA 5in 20 in = 6244 Solve for TA such that A1+A2=0 11.5 10 lb in  22.25in  .5 in 5  1  1 Case 1: = 8.857 10 lb in TA  TB1 60 lb  in 20 in  1200 lb  in AA21 8.857 105lb  1 in  1 T .01063 rad T( x )  1200  60 x   A

TA 120 lb in 20 2 1 1200 60x 1200 20 60 20  2 A1 dx T1200 lb  in  120 lb  in  JG JG B 0  BC  BC T =1080 lb in 12,000lb in2 B =- .01063rad  4 .5in  11.5 1062 lb in  1080lb in .5 in 2  max  5.50ksi BC  4 2 .5in 120lb in .25 in  max  4.89ksi AC  4 2 .25in

ENGR:2750 Mech. Def. Bodies The University of Iowa 11.10