1. Copyright Joseph Greene 2003 All Rights Reserved 1 CM 197 Mechanics of Materials Chap 17: Statically Indeterminate Beams Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials, 2 nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) CM 197

2. Copyright Joseph Greene 2003 All Rights Reserved 2 Design of Beams for Strength Topics –Introduction –Method of Superposition –Propped Cantilever Beams –Fixed Beams –Continuous Beams

3. Copyright Joseph Greene 2003 All Rights Reserved 3 Introduction Three types of beams have been considered previously –Simple beams –Overhanging beams –Cantilever beams All three are statically determinate –Reactions at the supports can be determined from equilibrium conditions. Statically INDETERMINATE –Beams for which the equilibrium equations alone are NOT sufficient to solve for reactions at the supports. –Additional equations based upon deflection conditions must be introduced to solve for external reactions. –There are more than three reaction components in the beam. Have redundant constraints. Number of redundant constraints is the degree of indeterminancy. –Figure 17-1 Propped cantilever beam has 1 redundant constraint, R B, Statically indeterminate (SI) to the first degree. Continuous beam has 2 redundant constraints, R B and R C, SI to the 2 nd Degree Fixed beam has 3 redundant constraints, R By, R Bx, M B, SI to the 3 rd Degree

4. Copyright Joseph Greene 2003 All Rights Reserved 4 Method of Superposition Solve statically INDETERMINATE (SI) problems- Like Chap 16. –SI 1 st degree (Propped cantilever beam) - one of the constraints is redundant and can be temporarily removed. Solve: –The reaction at the redundant constraint can be considered as an unknown load. –The slope or deflection at the removed constraint is set to a value compatible with the original conditions. –The shear forces and deflections are calculated as in Chapter 16 for statically determinate beams. Use Table 16-1. –SI 2 nd Degree (Continuous beam) – Two support elements are redundant and the reactions are treated as unknown loads. Solve with the slope or deflection at the removed constraint is set to a value compatible with the original conditions. Gives rise to two equations that can be used to solve for reactions and unknown supports. –SI 3 rd Degree (Fixed beam)- Three support elements are redundant and the reactions are treated as unknown loads. Solve with The slope or deflection at the removed constraint is set to a value compatible with the original conditions. Gives rise to three equations that can be used to solve for reactions and unknown supports.

5. Copyright Joseph Greene 2003 All Rights Reserved 5 Propped Cantilever Beams Have: –A fixed support on one end and a roller support on the other end. Fig 17-2a –A roller support at some intermediate point, Fig 17-2b. SI 1 st degree (Propped cantilever beam) - one of the constraints (Usually roller support) is redundant and can be temporarily removed. –Solve: Remove roller support and treat it as a unknown load. Solve for equilibrium loads based upon cantilever beam. Table 16-1 The shear forces and deflections are calculated as in Chapter 16 for statically determinate beams. –Example 17-1 –Example 17-2

6. Copyright Joseph Greene 2003 All Rights Reserved 6 Fixed Beams Have: –A fixed support on both ends and a zero slope and deflection at the ends. –Axial forces are absent as in Fig 17-3a and b Four unknown reaction elements and two static equilibrium conditions. SI 3 rd degree (Fixed beam) – three of the constraints (Usually end moments and no axial loads) are redundant and can be temporarily removed. –Solve: Assume moment actions on each end as redundant (removed) and release moment constraints on fixed end. Assume axial loads are absent. Beam becomes simply supported. Loads on simple beam consists of redundant moment reactions, M A and M B, and the given loads. Set slopes at ends = 0 and compatible with fixed end. Solve for redundant Moments M A and M B, based upon fixed beam. Table 16-1 The shear forces and deflections are calculated as in Chapter 16 for statically determinate beams. –Example 17-3 –Example 17-4 –Example 17-5

7. Copyright Joseph Greene 2003 All Rights Reserved 7 Continuous Beams Continues over more than one span, Fig 17-4. –Extra beam supports effectively reduces the maximum moment compared to a simple beam. –Method of super position can NOT be used to analyze beam. Can only be applied in a few simple cases. Other techniques are needed to solve this (Not covered in this class) –Following examples Example 17-6 Example 17-7 SI 2 nd degree (Continuous beam)