Question 20.7: Determine the maximum tensile and compressive forces in the ...

The vertical component of the force in the member EC resists the shear force in the panel DC. Therefore we construct the shear force influence line for the panel DC as shown in Fig. 20.18(b). From Eq. (20.19)

S_{\mathrm{K}}=R_{\mathrm{B}}={\frac{x_{1}}{L}}     (20.19)

the ordinate \mathrm{df}=2 \times 1.4 /(8 \times 1.4)=0.25 while from Eq. (20.20)

S_{\mathrm{K}}=-R_{\mathrm{A}}=-{\frac{L-x_{2}}{L}}      (20.20)

the ordinate \operatorname{cg}=(8 \times 1.4-3 \times 1.4) /(8 \times 1.4)=0.625.

Furthermore, we see that S_{\text {DC }} changes sign at the point j (Fig. 20.18(b)) where jd, from similar triangles, is 0.4 .

The member EC will be in compression when the shear force in the panel DC is positive and its maximum value will occur when the head of the load is at \mathrm{j}, thereby completely covering the length aj in the S_{\mathrm{DC}} influence line. Therefore

F_{\mathrm{EC}} \sin 45^{\circ}=S_{\mathrm{DC}}=2.5 \times \frac{1}{2} \times 3.2 \times 0.25

from which

F_{\mathrm{EC}}=1.41  \mathrm{kN} \quad \text { (compression) }

The force in the member EC will be tensile when the shear force in the panel DC is negative. Therefore to find the maximum tensile value of F_{\mathrm{EC}} we must position the load within the part \mathrm{jb} of the S_{\mathrm{DC}} influence line such that the maximum value of S_{\mathrm{DC}} occurs. Since the positive portion of the S_{\mathrm{DC}} influence line is triangular, we may use the criterion previously established for maximum bending moment. Thus the load per unit length over \mathrm{jb} must be equal to the load per unit length over \mathrm{jc} and the load per unit length over cb. In other words, c divides the load in the same ratio that it divides \mathrm{jb}, i.e. 1: 7. Therefore 0.5 \mathrm{~m} of the load is to the left of c, 3.5 \mathrm{~m} to the right. The ordinates under the extremities of the load in the S_{\mathrm{DC}} influence line are then both 0.3125 \mathrm{~m}. Hence the maximum negative shear force in the panel \mathrm{CD} is

which gives

S_{\mathrm{CD}}(\max -\mathrm{ve})=4.69  \mathrm{kN}

Then, since

\begin{aligned} F_{\mathrm{EC}} \sin 45^{\circ} & =S_{\mathrm{CD}} \\ F_{\mathrm{EC}} & =6.63  \mathrm{kN} \end{aligned}

which is the maximum tensile force in the member EC.