{"id":1325,"date":"2024-09-23T14:59:12","date_gmt":"2024-09-23T12:59:12","guid":{"rendered":"https:\/\/solveredu.com\/?p=1325"},"modified":"2026-01-27T23:48:56","modified_gmt":"2026-01-27T22:48:56","slug":"beams-with-hinges-example-solutions","status":"publish","type":"post","link":"https:\/\/solveredu.com\/en\/post\/obliczanie-belek-z-przegubem\/","title":{"rendered":"Beams with hinges - example solutions"},"content":{"rendered":"

In this entry:<\/p>\n\n\n\n

    \n
  1. An example solution for an beam with hinge.<\/a><\/li>\n\n\n\n
  2. An example solution for a multi-span beam.<\/a><\/li>\n<\/ol>\n\n\n\n

    Beams with hinge - Examples of problem solutions<\/h1>\n\n\n\n

    Below in the figure you will find a beam that we will solve. The beam has one inner joint<\/strong>.<\/p>\n\n\n\n

    \"Reactions<\/figure>\n\n\n\n

    We start the calculation of beams by checking the static determinability for the beam looks like this. For more on checking static determinability, see here<\/a>.<\/p>\n\n\n\n

    N=R-J-3<\/em>=4-1-3=0
    where:
    N - degree of static indeterminacy
    R =4 - number of support reactions
    J =1 - number of internal joints
    3 - the number of equilibrium equations. In static systems it is 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Since we have an internal hinge at point B in determining the degree of static determinability we substitute 1 for J. The result we get is zero so the beam is statically determinable. In the next step we will calculate the reactions from the equilibrium equations.<\/p>\n\n\n\n

    The beam diagram, all calculations and internal force diagrams are generated in my Beam Calculator<\/a>. You can create beam diagrams online and get a detailed solution for each statically determinate beam. <\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    We perform the calculation of reactions in the same way as for a straight beam. Because of the hinge, we write an additional equilibrium equation.<\/p>\n\n\n\n

    \"Reactions<\/figure>\n\n\n\n

    The sum of moments with respect to point B on the right side is zero. The joint allows us to write the equation for the bending moment for only one of the sides of the beam, either right or left. The choice is up to us. In this example, it is better to choose the right side.<\/p>\n\n\n\n

    We have the reactions calculated let's move on to determining the internal forces<\/a> In compartments.<\/p>\n\n\n\n

    In an articulated beam, we proceed in exactly the same way as in a straight beam. In the example in question, we have three compartments. You can find the determination of shear forces and bending moments in the figure below.<\/p>\n\n\n\n

    \"Results<\/figure>\n\n\n\n
    \"Results<\/figure>\n\n\n\n
    \"Results<\/figure>\n\n\n\n
    Note that the bending moment in the joint at point B is zero. This is because a joint, by definition, does not transmit a bending moment, so in tasks with a beam with a joint, a zero bending moment at the joint should always come out.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Once we have determined the value of shearing forces and bending moments at characteristic points, we can move on to drawing diagrams.<\/p>\n\n\n\n

    \"Results<\/figure>\n\n\n\n

    And at this point we will finish solving the example of a beam with a single internal joint.<\/p>\n\n\n\n

    Example of a solution for a multi-span beam<\/h2>\n\n\n\n

    As another example, we will solve a beam with two joints. The procedure is the same as for a beam with one joint. The difference will be when writing the equilibrium equations. In this case, we will have one more equation.<\/p>\n\n\n\n

    \"multi-beam,<\/figure>\n\n\n\n

    The static determinability check for such a beam is as follows.<\/p>\n\n\n\n

    N=R-J-3<\/em>=5-2-3=0
    where:
    N - degree of static indeterminacy
    R =5 - number of support reactions
    J =2 - number of internal joints
    3 - the number of equilibrium equations. In static systems it is 3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Since we have an internal hinge at point B and D in determining the degree of static determinability, we substitute \"2\" for J. The result we get is zero so the beam is statically determinable. In the next step we will calculate the reactions from the equilibrium equations.<\/p>\n\n\n\n

    \"Reactions<\/figure>\n\n\n\n

    We perform the calculation of reactions in the same way as for a straight beam. Because of the hinges, we write two additional equilibrium equations.<\/p>\n\n\n\n

    As for a beam with a single joint, we can choose the side relative to which we determine the bending moment in the joint. In this example, the better choice is the left side. Both for joint B and the joint at point D.<\/p>\n\n\n\n

    We have the reactions calculated let's move on to determining the internal forces<\/a> In compartments.<\/p>\n\n\n\n

    In an articulated beam, we proceed in exactly the same way as in a straight beam. In the example in question, we have as many as six compartments. You can find the determination of shear forces and bending moments in the figure below.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    Note that the bending moment in the joint at point B and in joint D is zero. This is because a joint, by definition, does not transmit a bending moment, so in tasks with a beam with a joint, a zero bending moment at the joint should always come out.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Finally, we will determine the courses of shearing forces and bending moments in the form of graphs.<\/p>\n\n\n\n

    \"hinged<\/figure>\n\n\n\n

    And this concludes the entry \"Calculation of beams with articulation\". Thank you for making it to the end.<\/p>","protected":false},"excerpt":{"rendered":"

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