For the beam and loading shown design the cross section of the beam, knowing that the grade of timber used has an allowable bending stress of 12 MPa. Don't consider the weight of the beam 3 kN/m 150 mm B IC -24 m -1.2 m-
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- An under-reinforced beam section given below. What could be the maximum design load (Pa) on the beam. Materials C30/37 and B420C. Ignore beams' own weight! Pa Pd H A 2 m 2 m, 2 m B As 1500 mm² 300 mm > Cross-section 470 mm 30 mmThe built-up beam wooden above is subjected to a horizontal shear force, V, as shown. Assume that the beam dimensions are drawn to scale. Also assume that the nails attaching the three pieces of the cross-section allow it to act as a single member. Which of the following best explains why we cannot analyze the shear stresses in this beam with the techniques covered in this class (at least, so far)? The shear force is not "transverse" O Wood cannot be loaded perpendicular to the grain The shear force is not acting along a line of symmetry of the cross-section We can only deal with vertical shear forces(1) Redo the following Calculate Mn ? = 448 14" 21" 3' wh Beam sections, to fé= 4 ksi Grade Co steel NWC M=110** for the all questions. (a) Computer the bending stresses in beam by using transformed-area for the all question the -area method.
- Determine the largest safe value of the load intensity w0 carried by the Ibeam if the working stresses are 3500 psi in bending and 260 psi in shear The wood beam has a square cross section. Find the smallest allowable cross-sectional dimensions if the working stresses are 8 MPa in bending and 1.0 MPa in shearProblem 1. The composite beam shown below carries a cantilevered load of 10 kN. The beam consists of one 30 x 124 mm plate and four 12 x 50 mm plates. They are pinned together at 120 mm intervals with round pins. The pin material has a shear strength of 159 MPa. Compute the minimum acceptable diameter for the pins. O O O O O O O -0 O 0- O 1000 mm Do O -120 mm (typ) O O O P = 10 KN 30 x 124 mm 12 x 50 mm (typ)2a) Justify the selection of a standard steel rolled section for the following Column from a bending moment diagram you determine that M is 200.7 MNmm at the support and the stress at that point in the beam is 175N/mm^2. Take into consideration an additional margin of safety as 20%.
- 2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80(NOS 2 – 13) PROBLEM 1. A timber beam shown in the figure is made up of Sasalit (50% stress grade), Use LRFD. All the given loads shown in the figure below are unfactored service loadings only, it does not include the weight of the beam itself. 8.) What is the maximum actual moment in KN-m? 9) What is the actual maximum bending stress in MPa? 10) What is the actual maximum shearing stress in MPa? 11) What is the moment of inertia of the beam in mm4 with respect to x-axis? 12) What is the moment of inertia of the beam in mm4 with respect to y-axis? 13) The allowable deflection for beams is L/240, what is the allowable deflection in mm.? PDL = 5 KN %3D 500 mm PLL = 3 KN Wp =6 KN/m WL 3 KN/m 800 mm R1 R2 2 m 1 m 1 mA laminated beam is composed of four (4) planks, each of 150 mm x 30 mm, glued together to form a section 150 mm x 150 mm. The allowable shear stress in the glue is 850 kPa, the allowable shear stress in the wood is 1050 KPA, and the allowable flexure stress in the wood is 10 MPa. Determine the maximum uniformly distributed load which can be carried by the beam on a 45 m simple span.
- A timber beam shown in the figure is made up of Sasalit (50% stress grade). Use LRFD. All thegiven loads shown in the figure below are unfactored service loadings only, it does not include the weight of the beam itself. R₁ WD = 6 KN/m W₁ = 3 KN/m 2 m PDL = 5 KN 1 m PLL = 3 KN 1 m 2) What is the unit weight of the given wood, in KN/m³? 4) What is the actual reaction R₁ in KN? R₂ 3) What is the unfactored weight of the beam itself in KN/m? 500 mm 800 mmRefer to the figure shown below. 350 Given are the following: fc 55 MPa fou 1,350 MPa Aps - 1.500 mnm2 fpy - 1,100 MPa Use bonded tendons as prestressed reinforcement. Determine the design monment capacity of the section. (kN-m)A 5 ft. long laminated wood beam is built up by gluing together three 2” x 6” boards to form asolid beam 6” x 6” cross section, as shown in the figure. Calculate the maximum allowable valueof P (lb.) such that both the (1) allowable shear stress in the glued joints of 100 psi and the (2)allowable bending stress of 1200 psi is not exceeded. (Disregard the weight of the beam).[Formula: Bending Stress: = (M)(y)/I; max = (M)(c)/I; Transverse Shear Stress: =(V)(Q)/(I*t); max = (V)(QNA)/(I*tNA)]