Related Concept Videos

Cranial Bones: Lateral View01:27

Cranial Bones: Lateral View

The lateral view of the cranium is dominated by temporal, sphenoid, and ethmoid bones.
The temporal bone forms the lower lateral side of the skull. The temporal bone is subdivided into several regions. The flattened upper portion is the squamous portion of the temporal bone. Below this area and projecting anteriorly is the zygomatic process of the temporal bone, which forms the posterior portion of the zygomatic arch. Posteriorly is the mastoid portion of the temporal bone. Projecting...
Method of Sections: Problem Solving I01:27

Method of Sections: Problem Solving I

Consider a symmetrical roof truss structure, composed of vertical, diagonal, and horizontal members. The length of each horizontal member is 4 m. The lengths of the vertical members FB and HD are 4 m, while the length of member GC is 6 m. The loads acting at joints F, G, and H are 2 kN, while those at joints A and E are 1 kN.
Method of Sections: Problem Solving II01:30

Method of Sections: Problem Solving II

Consider an arbitrary truss structure composed of diagonal, vertical, and horizontal members fixed to the wall. To calculate the force acting on members CB, GB, and GH, method of sections can be used. The loads and lengths of the horizontal and vertical members are known parameters, as shown in the figure.
Shear on the Horizontal Face of a Beam Element01:16

Shear on the Horizontal Face of a Beam Element

To understand shear on the flat side of a prismatic beam element, consider the vertical and horizontal shearing forces, and the normal forces, acting on the element. The element's upper (U) and lower (L) sections, which are divided by the beam's neutral axis, are examined. The equilibrium of these forces is determined by applying the equilibrium equation, which helps identify the horizontal shearing force. This force is directly related to the bending moments and the cross-section's first...
Horizontal Curve: Problem Solving01:03

Horizontal Curve: Problem Solving

A horizontal curve is characterized by its radius, intersection angle, and stationing of key points. In this case, the radius is 400 meters, and the angle of intersection is 30 degrees, with the station of the point of curvature (P.C.) at 0 + 150 meters. The goal is to determine the station values at the point of intersection (P.I.), point of tangency (P.T.), and midpoint of the curve, as well as the length of the long chord.The process begins with calculating the tangent distance (T) and the...