At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 12.3 × 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 8.9 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. O 80.1°F O 118.6°F O 150.7°F 132.9°F O 110.9°F B ↑ 2 in. 44 in. 0.04-in. gap

At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 12.3 × 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 8.9 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. O 80.1°F O 118.6°F O 150.7°F 132.9°F O 110.9°F B ↑ 2 in. 44 in. 0.04-in. gap

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A steel angle iron, loaded in compression is added to a structure in order to increase its rigidity.…

Q: At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is…

A: Initial temperature (Ti) = 60°F Bar-1: Width of bar1 = 3 in. Thickness of bar1 = 0.75 in. L1 = 32…

Q: For the two element rod shown, element (1) is steel with: E₁ = 210 GPa, A₁ = 1000 mm², L₁= 2 m, and…

A: Given A two element rod having temperature increase of 100 °c Required Stress in both rod…

Q: Rigid

Q: The typical Young's modulus of steel is A 275-310 GPa B 145 - 170 GPa c) 195 - 210 GPa D) 225 - 240…

Q: At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is…

A: Given data in question Temperature 60°F0.04 in gap2 barsCoefficient of thermal expansion 14.4×10-6…

Q: A 300 mm Long copper wire of uniform cross-section is pulled in tension so that a maximum tensile…

A: Given that :- Maximum tensile stress = σ = 270 MPa Length = L = 300 mm Modulus of elasticity = E =…

Q: Materials can have different masses but equal volumes. a. TRUE b. FALSE

A: Density : It is the ratio of mass and volume . Mass : it defines about how much matter an object…

Q: 38. What grade of steel is preferred for the fabrication of most structural shapes other than…

A: Given data : Structural shapes Find : Grade of steel and its yield and tensile stress Shapes like W…

Q: At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is…

A: Given Data Find the lowest temperature (T2) at which the two bars contact each other.

Q: Two uniform spheres are positioned as shown. Determine the gravitational force F which the titanium…

Q: Question No.2 Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion…

A: It is that point on stress-strain curve, till which stress is linearly varying with strain. As per…

Q: 160KN 个 160KN 6cm 6cm 40 cm

Concept explainers

Alloys

An alloy is an admixture of two or more metals or a metal and a non-metal. Since ancient times, man has been using alloys. Alloying helps to retain the properties of the parent matter, such as hardness, strength, toughness, ductility and is corrosion-resistant. Copper alloys such as brass have antimicrobial properties that help to fight against (Coronavirus disease 2019) COVID-19 and other hospital-acquired infections. Some common alloys are brass, bronze, solder steel, etc. Most alloys have a lower melting point than their constituents. This is due to the presence of a nonmetal as one of the constituents that behaves as an impurity and restricts the raise of melting point.

Question

At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v =
0.32; a = 12.3 × 106/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 8.9 x
10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine the lowest temperature at
which the two bars contact each other.
(1)
3 in.
32 in.
O 80.1°F
O 118.6°F
O 150.7°F
O 132.9°F
O 110.9°F
B
2 in.
44 in.
0.04-in. gap

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1: Given data

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

A copper rod with a diameter of 24 mm is subjected to tension within the elastic region. If themodulus of elasticity is 115 GPa and a Poisson’s ratio of 0.34. Determine the force that willproduce a reduction of 0.003 mm in the diameter.
1. The tensile strength vs. temperature curves of aluminum are plotted below. Explain the meaning of these two curves. 400 Ultimate strength 300 Yield strength 200 100 -200 -150 -100 -50 50 100 150 200 250 Temperature, °C Strength, kPa
A polymer cylinder (2) is clamped between rigid heads by two steel bolts (1). The steel [E = 29,000 ksi; a = 6.5×10-6/°F] bolts have a diameter of 0.50 in. The polymer [E = 380 ksi; a = 39.7x 10-6/°F] cylinder has an outside diameter of 7.0 in. and a wall thickness of 0.40 in. Assume a = 22 in. and b = 30 in. If the %3D temperature of this assembly changes by AT = 125°F, determine: (a) the normal stress in the polymer cylinder. (b) the normal strain in the polymer cylinder. (c) the normal strain in the steel bolts. |(1) (2) C а (1) b. Answer:
A component made of Aluminium 6061-T651 has an edge crack with length equal to 15 mm, as shown The yield stress of the material is oy = 280 MPa and its fracture toughness (Kic) ranges from 40 MN/m3/2 to 60 MN/m³/2 Consider now the same edge crack in a semi-infinite plate. Determine the critical stress to avoid fracture. P I I I I ← 15 mm P↓ 100 mm 30 mm
A uniform beam of weight Wis supported by the two rods, the lower ends of which are at the same level just before the beam is supported, as shown in figure. The ratio of the areas Aluminium (E = 70 GPa) 2 m 1.5 m W 3 m 3m Aaluminium Asteel Steel (E = 200 GPa) 0.5 m of the rods so that the beam will be horizontal after it is attached to the rods is
#2 will upvote to anyone who will answer
If it is constrained between two supports A and B and is stress-free at 20 ℃, what would be the stress in the two materials when it is heated to 70 ℃, For Steel: Es = 210 GPa, Coefficient of expansion = 12x10^6/℃. For Brass: Eb = 105 GPa, Coefficient of expansion = 19x10^-6/℃
A rigid bar, AB, is pinned at point B. Two other bars, AC and AD, are connected to point A to provide support for AB. We are investigating the behavior of this system under a specific load. Key Details: Bar properties: Material: Steel with Young's modulus (Esteel) of 200 GPa and coefficient of thermal expansion (a) of 11.7 x 10^-6 /°C. Cross-sectional area: AC: 1960 mm² AD: 1250 mm² Length: AC: √2 meters (square root of 2 meters) AD: 1 meter Inclination angles: AC: θ₁ (theta 1) = 30° from horizontal AD: θ₂ (theta 2) = 20° from horizontal Load on bar AB: Idealized linearly increasing load with a maximum value of 30 kN/m at point B. Problem Objectives: Determine the internal forces (axial forces) acting within bars AC and AD due to the applied load on AB. Calculate the displacement of point A in both horizontal (dA,x) and vertical (dA,y) directions.If the bar is cooled by 10°C, what would be the: Determine the internal forces (axial forces) acting within bars AC and AD…
Question 1 A 0.5 mm 250 mm 300 mm B Magnesium A = 2000 mm² E=45 GPa α = 26 X 10-6/°C Aluminum A = 800 mm² E = 75 GPa a = 23 x 10-6°C 5 pts At room temperature (20°C) a 0.5-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 120°C. (a) The normal stress in the aluminum rod (MPa) =
A mild steel flat of width 120 mm and thickness 10 mm is bent into an arc of a circle of radius 10 m by applying a pure moment 'M'. If E is 200 GPa, then magnitude of the pure moment 'M' will be