AW 12 x 65 column section shown in the figure is pinned at each end and has an additional support in the weak direction at a point 4.0 m from the top. Use X = 5.2 m K 1.00 for both directions.

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A: By Muller Breslau method, As we have to find the reaction at end A, hence Support at A will be…

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A: For calculation assume,

Q: 5. Find diameter y. All dimensions are in millimeters. 1 + 19.30 DIA 118.00 -३०

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A:

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Q: Compute Ix for the region shown. y 2 18 2-5--5- -2 -8 8. Dimensions in inches

A: The area is divided into the standard areas as shown below.

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Q: 7⁄8 inch+  5⁄32 inch

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A:

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A: Given: Specimen LL PL PI Cc Cu 1 53 24 29     2       1.05 95.52 3 27 16 11 3.06 28.34 4…

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A: Step 1: dear students yours questions is insufficient please full answer and missing the information…

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Q: 31⁄5 / 4⁄7

A: 31/5=6.2 4/7=0.571

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A: The given figure shows the different items, it is required to state whether the Multiview projection…

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A: given- a plane figure to find- M.O.I  about X-axis and centroidal axis.

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A: Understand force flow. Understand the math behind structural analysis (concepts like virtual work…

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Q: The column section shown above is reinforced with equal number of 20 mm bars с 1. on each face.…

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Q: How do I get the support forces

A: The x and y are the vertical and horizontal distances: x=12cos60=6 fty=6sin60=5.196 ft

Q: Determine centroidal coordinates of the volume.

A: x¯=∑χiVi∑Vi   =610×660×6102 + 12×610×660×(610+13×610)610×660 + 12×610×660   = 474.45 mm y¯=8132    =…

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A: The correct punctuation marks for this be like this "I don't like this," said Ali.

Q: 350 mm 250 mm 600 mm Member Magnitude,N X-component Y-Component 180 mm 360 mm G H TBD 240 mm 400 mm…

A: Given- A frame structure is given    To find- forces at joints G H E and Tbd

Q: A Derive the equation for the unit influence line for the reaction at the support B in the…

A: By Muller Breslau method, remove Reaction at support B and place a Load of P as shown in figure. By…

Types of Composite

Composite materials are formed by merging two components that have different chemical and physical properties. The merged output is termed as a composite, whose properties are completely different from the individual components. Composite materials also have better strength than the individual components.Composite materials usually have two main parts, reinforcement material, and matrix material. The matrix material acts as a binder and surrounds the fiber of another material that bears a much higher strength. This material is known as the reinforcement material. The matrix materials are usually high-strength additives resins, while the fibers of the reinforced material can be random, oriented, and in a mat format.Composite materials are also known as fiber-reinforced composite polymers because they are composed of the polymer matrix which is reinforced to laboratory-created fibers or naturally occurring fibers, such as glass fibers, carbon fibers, or aramid fibers. Under the application of external loads, the matrix material absorbs and transfers the loads to the fibers, these fibers in return provide good compressive strength, tensile strength, and stiffness to the fiber-reinforced composite materials.Depending upon the matrix and the reinforcement material, different properties can be induced to the fiber-reinforced composite materials, including good corrosion resistance properties. The most common matrix material is the resin while the glass fibers act as the reinforcement material. Strong fibers provide strength and stiffness to the composite materials while, a soft and flexible resin provides shape to the composite materials.