Sample Problems for Exam 1. Te saft below as lengt L, Torsional stiffness GJ and torque T is applied at point C, wic is at a distance of 0.6L from te left (point ). Use Castigliano teorem to Calculate te maximum twist angle. T C B Solution: Denote te reaction torque applied at point B as TB. Ten te torque in te segment BC is equal to T B, and te torque in segment C is equal to T+T B. Te Strain ( ) T + TB (0.6 L) TB (0.4 L) energ isu + GJ GJ We find T B b differentiating U wit respect to T B and setting te derivative to ero (because te twist angle at point B is ero) U ( T + TB )(0.6 L) TB (0.4 L) + 0 TB 0.6T TB GJ GJ We now substitute TB into te energ expression and differentiate wit respect to T to get te twist angle at C, wic is te maximum twist angle. ( ) ( ) T L T L T L 0.4 (0.6 ) 0.6 (0.4 ) 0.1 U + GJ GJ GJ U 0.4TL φc T GJ. tin aluminum seet is to be used to form a closed tin-walled section. f te total lengt of te wall contour is 100 cm, wat is te sape tat would acieve te igest torsional rigidit? Consider elliptical (including circular), rectangular, and equilateral triangular sapes. Solution: Torsional rigidit is 4 J (1) ds t n tis problem t is constant and same for all sapes. lso ds 100 cm fixed for all te sapes. Terefore sape of largest would give te igest rigidit. Rectangular of ( a + b) 100
b a ab a b 10 40 400 15 35 55 5 5 65 Equilateral triangular of 3a100 a a sin 60 481.15 elliptical section b a perimeter100 π ( a + b ) ( a b) / πab
a b 10 0.88 656 15 16.8 79 15.9 15.9 796 ab15.9 is a circular section and offers te igest rigidit. 3. Tis is a part of Problem 6.65. solid circular saft as a diameter of 00mm. second saft as te same elastic sear strengt but is ollow wit an inside diameter of 150mm. Bot safts are torqued to teir sear strengt b torques of equal magnitude. Tis ields an outer diameter of te ollow saft of 17.75mm. Find te ratio of te full plastic torques of te two safts for elastic-perfectl plastic material. π 3 Solution: For te solid saft TFP τ D wile for te ollow saft 1 3 3 ( ) 3 3 3 π D 00 TFP τ Do Di so te ratio is r 1.15 3 3 3 3 1 D D 17.75 150 o 4. Te stringer- web sections sown in figures are subject to searv 0, wilev 0. Find te bending stress in te stringers for te same bending moment. Wic section is most effective in bending? Solution: (solution is in a coordinate sstem wit x being te beam axis, and - being te coordinates in te plane of te cross section. i M (a) c Given, V 0, 0 M 0 Z 0 (smmetr) xial stress due to Moment M M σ For te section (a), + 4. Terefore
M σ (1) 4 (b) 0 0 (Section is smmetric wit respect to ) + + + 4. Ten, M M σ () 4 (c) Lets find te centroid of te section (origin of te sstem), i i + 0 + 0 + c 0 4 i c 0 So, & is at centroid and. i i i ( )( ) + (0)( ) + (0)( ) + ( )( ) i i 4 i i
From equation 4.9 (wit M 0) M M σ + σ M + 4 4 4 4 8 4 8 4 M + (3) Neutral xis for te section (c) and moment M can be found as M tanα 1 α 45. f we plot tis on te section, M 1 45 3 4 Magnitude of te stresses (absolute value of te stress) will be identical at points and 3. Stresses at 1 and 4 are ero since te neutral axis passes troug tem. Terefore it is sufficient to ceck points and 3 for te maximum magnitude of te stresses. xial stress at points ( 1) & (4) σ 0 xial stress at points M 1 ( ) & (3) σ (4) Comparing (1), () and (4) we can conclude tat: - Section (b) and (a) as same effectiveness for bending. - Section (c) is te least effective as teσ is te largest altoug same amount of material is used in all (a), (b) and (c). 5. Wat are te equations for te in-plane displacements in torsion of a circular cross section? How do we see tat te correspond to rigid bod rotation of te section? Wat is different about te displacements in a non-circular cross-section?
nswer: Te displacements are: u β θ v+ x β x θ w 0. Te correspond to ero strains in u v v u te plane of te cross-section ε 0, ε 0, γ x + 0. For a noncircular cross section, u and v are still te same, but w is not ero x x (warping). 6. Consider an equilateral triangle and a bending moment tat in te elastic ranges produces maximum tensile stresses at a vertex and maximum compressive stresses on te side opposite tis vertex. f te material is elastic-perfectl plastic, were is te neutral axis for te full plastic moment? nswer: t is parallel to te side wic ad te maximum compressive stresses, and it divides te triangle into two equal areas. Since te area of te triangular side is proportional to te square of te lengt of te sides, tese ave to be reduced b square root of 7. Wen are te Euler Bernoulli assumptions invalid for slender beams? nswer: Wen te Sear modulus is muc smaller tan oung s modulus