نقطة واحدة What will be the variation in BMD for the diagram? [Assume 10 KN A B .L = 2m] Square O Triangular O Trapezoidal O Rectangular O
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![نقطة واحدة
What will be the variation in
BMD for the diagram? [Assume
10 KN
A
B
.L = 2m]
Square O
Triangular O
Trapezoidal O
Rectangular O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5759cc3a-df43-405c-9e30-7da441056b6d%2F0ea3ae53-481b-43d6-bb55-68b8f0543ef4%2Fub4jw9j_processed.jpg&w=3840&q=75)
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- 2. Determine the centroid for the following figure: 2.5 E2.5"- 1.5" 3"y=82 show complete solutionsThe thin rectangular plate shown is uniformly deformed such that ɛ = +874 pɛ, ɛ, = -536 pɛ, and yry= +658 prad. Determine the normal strain ɛAC along diagonal AC of the plate. %3D D 72 mm A |B 108 mm O 744 µE O 447 µE 0 939 µE O 790 µɛ Ο 613 με
- Determine the degree of indeterminacy for each of the following structures:Given data: Cross sectional area, A 2 mm2; Weight, W = 12 N; Extension, 8 = 0.6 mm %3D 3 Modulus of elasticity, E 150 Gpa = 150×10° N/mm´ %3D To find: Maximum length or wire Solution: Let, l maximum length of the wire, Extension of the wire under its own %3D weight, We 12xl 0.6 0.6 = 0.02x10-3l; l = 30000mm 30m %3D %3D %3D 2AE 2x2x(150x10) 0.02x103 Result: Maximum length of wire = 30 m Problem 15. A copper alloy wire of 1.5 mm diameter and 30 m long is hanging freely from a tower. When will be its elongation due to self weight? Take specific of the copper and its modulus of elasticity as 89.2 kN/m³and 90 Gpa respectively. Given data: 30x10 89.2 kN/m°= 89.2×10° kN/mm = 89.2×10°N/mm³ 90 Gpa = 90 x 10° N/mm². Diameter, d 1.5 mm; Length, L =30m mm 3 -6 Specific weight, w = Modulus of elasticity, E To find: Elongation of the wire. 8 x(d)* 품지(1.5)°-1 Solution: Cross-sectional area of the wire, A = %3Dfinal ans is 0.37
- Steel bars and an aluminum bar support the rigid slab as a platform as shown below. A gap A= 0.26 mm %3D exists between the aluminum bar and the slab. Neglecting the weight of the slab a (/degree celsius) 11.6x10* 23.1x10* A (mm²) E (GPa) Steel bar Aluminum bar 350 200 400 70 a) Which temperature increase AT is required to just close the spacing at the middle? (Note that the temperature increase will subject all bars) b) What are the stresses in the steel bars and aluminum bar for AT = 90 degree celsius in MPa? 1000 mm Steel Aluminum SteelFN=130 MN=55Steel railroad rails 10 m long are laid with end-to-end clearance of 5 mm at a temperature of 18°C. At what temperature will the rails just come in contact. Use a = 11.7 x 10-6/°C and E = 200 GPa. %3D %3D