wood

Solution to Problem 581 | Design for Flexure and Shear

581 Five Planks Laminated Beam

Problem 581

A laminated beam is composed of five planks, each 6 in. by 2 in., glued together to form a section 6 in. wide by 10 in. high. The allowable shear stress in the glue is 90 psi, the allowable shear stress in the wood is 120 psi, and the allowable flexural stress in the wood is 1200 psi. Determine the maximum uniformly distributed load that can be carried by the beam on a 6-ft simple span.

 

Solution to Problem 572 | Horizontal Shearing Stress

572 Wooden T-beam Made From Rectangular Beams

Problem 572

The T section shown in Fig. P-572 is the cross-section of a beam formed by joining two rectangular pieces of wood together. The beam is subjected to a maximum shearing force of 60 kN. Show that the NA is 34 mm from the top and the INA = 10.57 × 106 mm4. Using these values, determine the shearing stress (a) at the neutral axis and (b) at the junction between the two pieces of wood.

 

18 - 20 Rectangular beam in maxima and minima problems

DiffCalc 003 Rectangular beam from a circular log

Problem 18

The strength of a rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the largest beam that can be cut from a log of given size.

 

Solution to Problem 527 | Flexure Formula

Problem 527

In Prob. 526, if the load on the overhang is 600 lb/ft and the overhang is x ft long, find the maximum values of P and x that can be used simultaneously.

 

Solution to Problem 526 | Flexure Formula

526-527 Rectangular wood beam

Problem 526

A wood beam 6 in wide by 12 in deep is loaded as shown in Fig. P-526. If the maximum flexural stress is 1200 psi, find the maximum values of wo and P which can be applied simultaneously?

 

Solution to Problem 129 Bearing Stress

129 bolted timbers

Strength of Materials 4th Edition by Pytel and Singer
Problem 129 page 21

Given:
Diameter of bolt = 7/8 inch
Diameter at the root of the thread (bolt) = 0.731 inch
Inside diameter of washer = 9/8 inch
Tensile stress in the nut = 18 ksi
Bearing stress = 800 psi

Required:
Shearing stress in the head of the bolt
Shearing stress in threads of the bolt
Outside diameter of the washer

Solution to Problem 123 Shear Stress

123 Forces parallel and perpendicular to wood grain

Strength of Materials 4th Edition by Pytel and Singer
Problem 123 page 18

Given:
Cross-section of wood = 50 mm by 100 mm
Maximum allowable compressive stress in wood = 20 MN/m2
Maximum allowable shear stress parallel to the grain in wood = 5 MN/m2
Inclination of the grain from the horizontal = 20 degree

Required: The axial force P that can be safely applied to the block

Solution to Problem 122 Shear Stress

122 Normal and shear components of the resultant on arbitrary section

Strength of Materials 4th Edition by Pytel and Singer
Problem 122 page 18

Given:
Width of wood = w
Thickness of wood = t
Angle of Inclination of glued joint = \theta
Cross sectional area = A

 

Required: Show that shearing stress on glued joint \tau = P \sin 2 \theta \, / \, 2A

Solution to Problem 110 Normal Stress

110 Wooden post in concrete footing

Strength of Materials 4th Edition by Pytel and Singer
Problem 110 page 13

Given:
Size of steel bearing plate = 12-inches square
Size of concrete footing = 12-inches square
Size of wooden post = 8-inches diameter
Maximum allowable stress for wood = 1800 psi
Maximum allowable stress for concrete = 650 psi

Required: Maximum safe value of load P

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