(a)
Write the values of modulus of resilience and modulus of toughness for titanium.
The range of modulus of resilience for titanium is
The range of modulus of toughness for titanium is
(a)

Explanation of Solution
Refer Table 17.4 “The strength of Selected Materials” in the textbook.
The range for Yield strength of titanium alloys
The range for Ultimate strength of titanium alloy
Refer Table 17.2 “Modulus of Elasticity and Shear Modulus of Selected Materials” in the textbook.
The modulus of elasticity of titanium alloy is
The elongation of titanium alloy is 23%.
Find the range of modulus of resilience
Substitute the minimum and maximum values;
Therefore, the range of modulus of resilience is
Find the range of modulus of toughness
Substitute minimum and maximum values;
Therefore, the range of modulus of toughness is
(b)
Write the values of modulus of resilience and modulus of toughness for steel.
The range of modulus of resilience for steel is
The range of modulus of toughness for steel is
(b)

Explanation of Solution
Refer Table 17.4 “The strength of Selected Materials” in the textbook.
The range for Yield strength of steel
The range for Ultimate strength of steel
Refer Table 17.2 “Modulus of Elasticity and Shear Modulus of Selected Materials” in the textbook.
The range for modulus of elasticity of steel (E) is
The elongation of steel is 15%.
Modulus of resilience:
Substitute the minimum and maximum values in Equation (1).
Therefore, the range of modulus of resilience is
Modulus of toughness;
Substitute minimum and maximum values in Equation (2);
Therefore, the range of modulus of toughness is
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Chapter 17 Solutions
EBK ENGINEERING FUNDAMENTALS: AN INTROD
- No chatgpt plsarrow_forward11. The prestressed T beam shown below is pretensioned using low relaxation stress-relieved Grade 270 strands. The steel area Aps = 2.5 in². The tensile strength is fpu = 270 ksi, and the concrete compressive strength is fr = 6000 psi. (a) Calculate the nominal moment strength Mn with hr = 6 in. 22" 15" T hf (b) Since this beam is a T-beam, the nominal moment strength M₁ increases with a thicker hf. However, M, stops increasing if he reaches a value. Determine the minimum thickness hy that can achieve the maximum nominal moment strength Mr. Also, calculate the corresponding maximum nominal moment strength Mn with the computed hf.arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward
- 10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward7. Match the given strand profiles with the corresponding loading conditions for a prestressed concrete (PSC) beam. Strand profile (b) (d) (c) (a) Ꮎ Load on a beamarrow_forward4. For serviceability considerations, the effective moment of inertia (Ie) is calculated using the following formula: le 1 - 1cr ((2/3) Mcr) Ma 2 - وا ≥ Note that the upper bound was previously set as Iut in the earlier ACI equation. (a) Arrange the following moment of inertia values in ascending order (from smallest to largest): le, Ier, Ig and lut (b) Mer is the cracking moment. Choose the cross-section that should be used to compute Mcr. NA. h 5. Identify and circle the figure that represents the scenario in which the torsional effect is permitted to be reduced according to the ACI code provisions. (3 pts) mt mi B (b)arrow_forward
- I will rate, thanksarrow_forward. 9. A reinforced concrete beam is subjected to V/ = 40 kips and Tu/ = 12 ft kips at the critical section. Given conditions: ⚫ Longitudinal reinforcements use No. 8 grade 60 steel with an effective depth d = 20 in. For shear capacity, V = 18 kips and V₂ = 22 kips • For transverse reinforcements, use No. 3 bars with grade 60. • The effective torsional area of A. = 150 in². • Crack angle = 45° ⚫ The minimum stirrup spacing is Smin = 4" and the maximum stirrup spacing is Smax = Find the required stirrup spacing at the critical section. 8".arrow_forward3. The beam shown on the right uses three No. 8 bars made of Grade 60 steel as longitudinal reinforcement. The allowable maximum center-to-center spacing of the longitudinal rebars has been determined to be 10 inches. Now assume that Grade 80 steel will be used instead. Determine whether the beam satisfies the rebar spacing requirements according to the ACI Code. Additional assumptions: • Estimate fs = fy • 20" Clear cover: ? 12" Clear side cover: 1.5" The clear cover depth cc and the clear side cover remain unchanged, regardless of the change in material.arrow_forward
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