In the steel structure shown, a 6-mm-diameter pin is used at C and 10- mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD (18 mm width and 6 mm thickness) with length of 150 mm. Assume E of the link BD is 200 GPa. P = 2 kN Determine Factor of Safety with respect to shear in the pin D if the ultimate shear is 150 MPa (2 decimals). Front view 6 mm 18 mm B IB Side view 120 mm 160 mm Top view 11:0

In the steel structure shown, a 6-mm-diameter pin is used at C and 10- mm-diameter pins are used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD (18 mm width and 6 mm thickness) with length of 150 mm. Assume E of the link BD is 200 GPa. P = 2 kN Determine Factor of Safety with respect to shear in the pin D if the ultimate shear is 150 MPa (2 decimals). Front view 6 mm 18 mm B IB Side view 120 mm 160 mm Top view 11:0

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.2.1P

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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.

Question

In the steel structure shown, a 6-mm-diameter pin is used at C and 10-
mm-diameter pins are used at B and D. The ultimate shearing stress is
150 MPa at all connections, and the ultimate normal stress is 400
MPa in link BD (18 mm width and 6 mm thickness) with length of 150
mm. Assume E of the link BD is 200 GPa. P = 2 kN
Determine Factor of Safety with respect to shear in the pin D if the
ultimate shear is 150 MPa (2 decimals).
Front view
6 mm
18 mm
B
B
Side view
120 mm
160 mm
Top view
11:0

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